Calculus 2

Credits: The below is basically all based on the notes by my professor for Calc 2 with some theorems and stuff copied directly.

Appendix E: Summation

Let m and n be positive integers with $n \geq m$

$$ a_m, a_{m + 1}, a_{m + 2}, … , a_n \in \mathbb{R} $$

Summation Notation

$$ \sum_{i=m}^{n} a_i = a_m + a_{m + 1} + a_m + a_{m + 1} + a_{m + 2} + … + a_n$$

Examples

$$ \sum_{t = 1}^{3} \frac{1}{t} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} = \frac{11}{6} $$

Properties of Summation

Let m and n be positive integers, where $n \geq m$. Let $c \in \mathbb{R}$.

Let $a_m, a_{m+1}, a_{m + 2}, … a_n \in \mathbb{R}$
Let $b_m, b_{m+1}, b_{m + 2}, … b_n \in \mathbb{R}$

$$ \sum_{i = m}^{n} (ca_i) = c \sum_{i = m}^{n} (a_i)$$
$$ \sum_{i = m}^{n} (a_i + b_i) = \sum_{i = m}^{n} (a_i) + \sum_{i = m}^{n} (b_i) $$
$$ \sum_{i = m}^{n} (a_i - b_i) = \sum_{i = m}^{n} (a_i) - \sum_{i = m}^{n} (b_i) $$
$$\sum_{i=1}^{n} c = c \cdot n$$
$$\sum_{i=m}^{n} c = c \cdot (n - m + 1)$$
m, n, c, and k are all constants.
$$ \sum_{i=m}^{n} c = \sum_{i=m}^{k-1} + \sum_{k}^{n}$$

Common Summation Formulas

1.

$$ \sum_{i = 1}^{n} i = \frac{n(n + 1)}{2}$$

Proof:
$$ \text{First note that }$$
$$ 2 \sum_{i = 1}^{n} = \sum_{i = 1}^{n} + \sum_{i = 1}^{n} $$
$$ = (1 + 2 + 3 + 4 + … (n - 1) + n) + (n + (n - 1) + (n - 2) + 4 + 3 + 2 + 1) $$
$$ = \underbrace{(n + 1) + (n + 1) + (n + 1) … (n + 1)}_{n \text{ times}} $$

2. Telescoping Sum Formula

$$ \sum_{i = 1}^{n} (a_{i + 1} - a_i) $$
$$ = (a_2 - a_1) + (a_3 - a_2) + (a_4 - a_3) … (a_n - a_{n - 1}) + (a_{n + 1} - a_n) $$
$$ = - a_1 + a_{n + 1} = a_{n + 1} - a_{1} $$

Similarly:
$$ \sum_{i = 1}^{n} (a_{i} - a_{i - 1}) = a_{n} - a_0$$

3.

$$ \sum_{i = 1}^{n} i^2 = \frac{n(n + 1)(2n + 1)}{6} $$

Proof :
Use telescoping formula to get the following:
$$ \sum_{i = 1}^{n} ((i + 1)^3 - i^3) = (n + 1)^3 - 1^3 = n^3 + 3n^2 + 3n $$
But also:
$$ \sum_{i = 1}^{n} ((i + 1)^3 - i^3) = \sum_{i = 1}^{n} (i^3 + 3i^2 + 3i + 1 - i^3) = \sum_{i = 1}^{n} (3i^2 + 3i + 1)$$
$$ = \sum_{i = 1}^{n} (3i^2) + \sum_{i = 1}^{n} (3i) + \sum_{i = 1}^{n} 1$$
$$ = 3 \sum_{i = 1}^{n} (i^2) + 3 \sum_{i = 1}^{n} (i) + n $$
$$ = 3 \sum_{i = 1}^{n} (i^2) + 3 \frac{n(n + 1)}{2} + n $$

$$ \iff 3 \sum_{i = 1}^n (i^2) = (n^3 + 3n^2 + 3n) - n - \frac{3 n^2 + 3n}{2}$$
$$ \iff 3 \sum_{i = 1}^n (i^2) = \frac{(2\cdot n^3 + 2 \cdot 3n^2 + 2\cdot 3n)}{2} - \frac{2n}{2} - \frac{3 n^2 + 3n}{2}$$
$$ \iff 3 \sum_{i = 1}^n (i^2) = \frac{(2n^3 + 6n^2 + 6n)}{2} - \frac{2n}{2} - \frac{3 n^2 + 3n}{2}$$
$$ \iff 3 \sum_{i = 1}^n (i^2) = \frac{(2n^3 + 3n^2 + n)}{2} $$
$$ \iff \sum_{i = 1}^n (i^2) = \frac{n(2n+1)(n+1)}{6} $$

4.

$$ \sum_{i = 1}^{n} i^3 = \left[\frac{n(n + 1)}{2}\right]^2$$

Examples Using Sum Formula

Evaluate
$$ \sum_{i = 1}^{5} (4i + 2)$$
$$ \sum_{i = 1}^{5} (4i + 2) = \sum_{i=1}^{5}(4i) + \sum_{i=1}^{5}(2)$$
$$ = 4 \sum_{i = 1}^{5} i + \sum_{i = 1}^{5} 2 = 4 \frac{5 (6)}{2} + 5(2) $$
Evaluate using telescoping sum formula
$$ \sum_{i = 1}^{5} (2^i - 2^{i - 1}) = 2^5 - 2^0$$
More Complicated Example
$$ \lim_{n\to\infty} \sum_{i = 1}^{n} \frac{8}{n} \frac{i^3}{n^3} $$
$$ = \lim_{n\to\infty} \sum_{i = 1}^{n} \frac{8i^3}{n^4} $$
$$ = \lim_{n\to\infty} \frac{8}{n^4} \sum_{i = 1}^{n} i^3 $$
$$ = \lim_{n\to\infty} \frac{8}{n^4} \frac{n^2 (n+1)^2}{4} $$
$$ = \lim_{n\to\infty} \frac{2 (n^2 + 2n + 1)}{n^2} $$

5.1: Sums

Riemann Sums

$$ x_i = a + i \Delta x$$
$$ c_i \in [x_{i-1}, x_i]$$
$$ C_n = {c_1, c_2, c_3, … c_n }$$

Divide [a, b] into n equal subintervals with endpoints a = $x_0 < x_1 < x_2 < … < x_n = b$. Each subinterval has length $\frac{b-a}{n} = \Delta x$

$$ R_n (f, C_n) = \sum_{i = 1}^{n} f(c_i) \Delta x$$

If $lim_{n\to\infty} (R_n(f, C_n))$ exists and is the same regardless of how $C_n$ is chosen, $f$ is integrable on [a, b] and we define the following:
$$ \int_{a}^{b} f(x) , dx = \lim_{n\to\infty} (R_n (f, C_n))$$

Special Riemann Sums

The n-th Right-Endpoint Sum

Choose $c_i$ to be the right-endpoint
Set $c_i = x_i$
$$ R_n = \sum_{i=1}^{n} f(x_i) \Delta x$$

The n-th Left-Endpoint Sum

Choose $c_i$ to be the left-endpoint
Set $c_i = x_{i - 1}$
$$ R_n = \sum_{i=1}^{n} f(x_{i-1}) \Delta x$$

The n-th Left-Endpoint Sum

Choose $c_i$ to be the midpoint
Set $c_i = \frac{x_{i - 1} + x{i}}{2}$
$$ R_n = \sum_{i=1}^{n} f\left(\frac{x_{i - 1} + x_{i}}{2}\right) \Delta x$$

Examples

Example 1

$f(x) = cos\left(\frac{x}{4}\right) + 2$
Interval $[0, 8\pi]$
Let n = 4

$$ \Delta x = \frac{8\pi - 0}{4} = 2\pi$$
$$x_0 = 0, \ x_1 = 2\pi, \ x_2 = 4\pi, \ x_3 = 6\pi,\ x_4 = 8\pi $$

$$ R_4 = \sum_{i=1}^{4} f(x_i) \Delta x = \sum_{i=1}^{4} f(x_i) 2\pi = 2\pi \sum_{i=1}^{4} f(x_i) = 2\pi [f(x_1) + f(x_2) + f(x_3) + f(x_4)]$$

$$ L_4 = \sum_{i=1}^{4} f(x_{i-1}) \Delta x = \sum_{i=1}^{4} f(x_{i-1}) 2\pi = 2\pi \sum_{i=1}^{4} f(x_{i-1}) = 2\pi [f(x_0) + f(x_1) + f(x_2) + f(x_3)]$$

Example 2

$f(x) = 2x + 1$
$$ n = 5$$
$$ \Delta x = \frac{b - a}{n} = \frac{10 - 0}{5} = 2 $$
$$ x_0 = 0,\ x_1 = 2,\ x_2 = 4,\ x_3 = 6,\ x_4 = 8,\ x_5 = 10$$
Note: $m_i$ is the midpoint
$$ m_1 = 1,\ m_2 = 3,\ m_3 = 5,\ m_4 = 7,\ m_5 = 9$$

$$ M_5 = \sum_{i=1}^{5} f\left(\frac{x_{i-1} + x_i}{2}\right) \Delta x = \sum_{i=1}^{5} f(m_i) \Delta x = \sum_{i = 1}^{5} f(m_i) 2 $$

Remarks

A = area under the graph of $f$ on [a, b]

If $f$ is increasing, then $R_n \ge A$
If $f$ is increasing, then $L_n \le A$

If $f$ is decreasing, you get the opposite of the above

More Examples

Example 1

Find a region where area is given by
$$ \lim_{n\to\infty} \left(\sum_{i=1}^{n}\left[\left(\frac{5}{n}\right)\left(3 + i \left(\frac{5}{n}\right)\right)^3\right] \right) $$

$$ \Delta x= \left(\frac{5}{n}\right) $$
$$ a = 3 $$
$$ f(x) = x^3$$
$$ \frac{b - a}{n} = \frac{5}{n} \text{ and } a = 3 \text{ , so } \frac{b - 3}{n} = \frac{5}{n} \implies b = 8 $$
$$ [a, b] = [3, 8 ] $$

5.3: Fundamental Theorem of Calculus Pt 2

Let $f$ be continuous on $[a, b]$. Let $F$ be an antiderivative of $f$ on $[a, b]$ (i.e. $F(x) = f(x)\ \forall x \in [a, b] $). Then the following is true:

$$ \int_{a}^{b} f(x) \ dx = F(b) - F(a) $$

Proof: f(x) is continuous on $[a, b]$. Use riemann sums and get a telescoping sum. Use Mean Value Theorem

Fundamental Theorem of Calculus Pt 1

Let f be continuous on $[a, b]$

5.5: The Substitution Rule

If f is even, then
$$\int_{-b}^{b} f(x) \ dx = 2 \int_{0}^{b} f(x) \ dx$$

If f is odd, then
$$\int_{-b}^{b} f(x) \ dx = 0$$

6.1: Area between Two Curves

$$ A = \int_{a}^{b} (g(x) - f(x))\ dx $$
where $\forall x \ g(x) > f(x)$

6.2: Volume of Solid

$$ V = \int_{a}^{b} A(x) \ dx $$
where $A(x)$ is the area of cross section at x.

For the volume of a solid of revolution (disk method):
$$ V = \int_{a}^{b} f(x)^{2} \pi \ dx$$

A more general formula for a solid of revolution (washer method):
$$ V = \int_a^b \pi (R(x)^2 - r(x)^2) \ dx $$
where $R(x)$ is the distance between the axis of revolution and the upper curve (farther from axis of revolution) and $r(x)$ is the distance between the axis of revolution and the curve that’s closer to the axis of revolution.

6.5: Average Value of Function

Average value from a to b:
$$ f_{avg} = \frac{1}{b - a} \int_{a}^{b} f(x) \ dx $$

7.1: Integration by Parts

Formula

Let u and v be differentiable functions
$$ \int u(x) v’(x) \ dx = u(x) \ v(x) - \int v(x) u’(x) \ dx$$
or
$$ \int u dv = uv - \int v\ du$$

Derivation

Use Product rule
$$ \frac{d}{dx} \left(u(x) \cdot v(x)\right) = u(x) \cdot v’(x) + u’(x) \cdot v(x)$$
$$ u(x) \cdot v(x) = \int u(x) \cdot v’(x) + \int u’(x) \cdot v(x)$$

Notes on using Integration by Parts

Note: Integration by parts is useful when $\int v du$ is easier to compute than the $\int u dv$. So choosing u is very important. You want du to be “simpler” than u.

How to choose u in Integration by Parts (Ranked from best to worst)

Logarithmic Functions
Inverse Trig Functions
Polynomials
Exponential Functions (Just leads to the same exponential and a constant)
Trigonometric Functions (Often just leads to another trig function)

Examples

Example 1

$$ u = ln(x), \ du = 1 \cdot dx$$
$$ du = \frac{1}{x} dx, \ v = x$$
$$ \int_{1}^{e^3} ln(x) \cdot 1 \ dx = \left. uv \right\vert_{1}^{e^3} - \int_{1}^{e^3} v du$$
$$ = \left. x ln(x) \right\vert_{1}^{e^3} - \int_{1}^{e^3} x \left(\frac{1}{x}\right) dx$$

7.2: Trigonometric Functions Integration Techniques

Integrating Odd powers of Sin/Cos

If $m$ is odd, then $m = 2k + 1, k \in \mathbb{Z}$
$$ \int \sin^m (x) \cos^n(x) dx = \int \sin ^{2k+1}(x) \cos^n(x)dx $$
$$ = \int \sin^{2k}(x)\cos^n(x)\sin(x)dx = \int (\sin^2(x))^k \cos^n(x) \sin(x)dx $$
$$ = \int (1 - \cos^2(x))\cos^n(x)\sin(x)$$

Use u substitution to solve

Integrating Even Powers of Sin/Cos

Helpful Identities and Formulas

$$ \int \sin(qx)\ dx = - \frac{\cos(qx)}{q} + C$$
$$ \int \cos(qx)\ dx = \frac{\sin(qx)}{q} + C$$
where q is a integer

$$ \sin^2(x) = \frac{1 - \cos(2x)}{2}$$
$$ \cos^2(x) = \frac{1 + \cos(2x)}{2}$$

Solving Technique

If $m \leq n$ and they’re both even:
$$ \int sin^m(x) cos^n(x) dx = \int \sin^{2k}(x)\cos^n(x)dx = \int (\sin^2(x))^k cos^n(x) dx$$
$$ = \int (1 - cos^2 (x))^k cos^n(x)dx$$
After this just use the half-angle formulas and integrate

if $n \leq m$ then just make everything in terms of $sin$ and do the same thing.

Integral of Tan

$$ \int \tan(x)dx = \ln(|\sec(x)|) + C$$
Derivation: Use u substitution and use the fact that $tan(x) = \frac{\sin(x)}{\cos(x)}$

Integral of Sec

$$ \int \sec(x) dx = \ln(|\sec(x) + \tan(x)|) + C$$

Integrating Odd Powers of Tan and Sec

Example:
$$ \int \tan^5 (x) \sec^3(x) \ dx$$
$$ \int \tan^4 (x) \sec^2(x) \sec(x) \tan(x) \ dx$$
$$ \int (\sec^2 (x) - 1)^2 \sec^2(x) \sec(x) \tan(x) \ dx$$
$$ u = \sec(x), \ du = \sec(x) \tan(x) dx $$
$$ = \int (u^2 - 1)^2 u^2 du $$
Solve now

Integrating Even Powers of Tan and Sec

$$ \int sec^m (x) tan^n(x) dx = \int sec^{2j} (x) tan^n(x) dx$$
$$ = \int sec^{2j - 2}(x) tan^n(x)sec^2(x)dx = \int (sec^2(x)^{j-1} tan^n(x) sec^2(x) dx)$$
$$ = \int (tan^2(x) + 1)^{j-1} tan^n(x) sec^2(x) dx$$
$$ u = tan(x), \ du = sec^2(x)dx$$
$$ = \int (u^2 + 1)^{j-1} u^n du$$

Integral of Tan

$$ \int cotan(x)dx = \ln(|\sin(x)|) + C$$
Derivation: Use u substitution and use the fact that $cotan(x) = \frac{\cos(x)}{\sin(x)}$

Integral of Sec

$$ \int \csc(x) dx = \ln(|\csc(x) - cotan(x)|) + C$$

Integrating Powers of Cotan and Cosec

Same as tan and sec (above) except use the following
$$ \csc^2(x) = cotan^2(x) + 1$$
$$ \frac{d(cotan(x))}{dx} = - \csc^2(x)$$
$$ \frac{d(\csc(x))}{dx} = - \csc(x) cotan(x)$$

7.4 Integrating with Partial Fractions

If we want to integrate the following, where $P(x)$ and $Q(x)$ are polynomials,
$$ \int \frac{P(x)}{Q(x)} $$

If $degree(P(x)) \geq deg(Q(x))$ then just do long division to simplify.

Case I: Q(x) is the Product of Distinct Linear Factors

Once we have $deg(Q(x)) > deg(P(x))$, then we can say the following:

$$ \frac{P(x)}{Q(x)} = \sum_{i=1}^{n} \frac{A_i}{a_i x - b_i} $$
where
$A_1, \ A_2, \ A_3,\ … A_n$ are constants.

Then we can integrate more easily since the following is true
$$\int \frac{A_i}{a_i x - b_i} = \frac{A_i}{a_i} ln(|a_i x - b_i|) + C $$

Case II: Q(x) has Repeated Linear Factors

$$ \frac{P(x)}{Q(x)} = \sum_{i=1}^{n} \frac{A_i}{(a_i x - b_i)^n} $$
where $A_1, \ A_2, \ A_3,\ … A_n$ are constants.

Example:

$$ \int \frac{1}{x^3 (x - 1)} \ dx$$
$$ \frac{1}{x^3 (x - 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-1}$$
$$ x^3 (x-1) \cdot \frac{1}{x^3 (x - 1)} = \left(\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{D}{x-1}\right) \cdot (x^3(x -1))$$
$$ 1 = Ax^2(x - 1) + Bx(x -1) + C(x-1) + Dx^3$$
$$ 1 = (A+D)x^3 + (B-A)x^2 + (C-B)x + (-C)$$
$$ 0x^3 + 0x^2 + 0x + 1 = (A+D)x^3 + (B-A)x^2 + (C-B)x + (-C)$$

$$ 0 = A+D$$
$$ 0 = B-A$$
$$ 0 = C-B$$
$$ 1 = -C $$
Solve for the other stuff and integrate

Case III: Q(x) with a Prime Quadratic Factor

$Q(x)$ has a prime quadratic factor in
$$\int \frac{P(x)}{Q(x)} \ dx $$

The prime quadratic factor $ax^2 + bx + c$ of $Q(x)$ will contribute a partial fraction decomposition of
$\frac{Dx+E}{ax^2 + bx + c}$

Example

$$ \frac{3x^2 + x}{x(x-1)(x^2 + 1)}$$
Here $x^2 +1$ is the irreducible prime quadratic factor.
$$ \frac{3x^2 + x}{x(x-1)(x^2 + 1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{Cx + D}{x^2 + 1}$$

Solve for A, B, and D.

Case IV: Q(x) has Repeated Prime Quadratic Factors

$Q(x)$ has repeated prime quadratic factors in
$$ \int \frac{P(x)}{Q(x)} \ dx $$

The prime quadratic factor $\left(ax^2 + bx + c\right)^n$ of $Q(x)$ will contribute a partial fraction decomposition of
$$ \frac{D_1x+E_1}{ax^2 + bx + c} + \frac{D_2x+E_2}{ax^2 + bx + c} + \frac{D_3x+E_3}{ax^2 + bx + c} + \frac{D_nx+E_n}{ax^2 + bx + c}$$

Example

$$ \frac{3x-1}{x^3(x^2 + 9)^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3} + \frac{Dx + E}{x^2 + 9} + \frac{Fx + G}{(x^2 + 9)^2}$$

7.8: Improper Integrals

Type I: Infinite Integrals

If the following is true:
$$ \forall t \ge a, \ \int_{a}^{t} f(x)\ dx$$
then,
$$ \int_{a}^{\infty} f(x)\ dx = \lim\limits_{t\to\infty} \int_{a}^{t} f(x)\ dx$$
If it converges, then the limit exists.
If it diverges, then the limit does not exist and the limit is infinity or -infinity.

If both of the improper integrals on the right side of the equation converge, the following is true:
$$ \int_{-\infty}^{\infty} g(x)\ dx = \int_{-\infty}^{a} g(x) \ dx + \int_{a}^{\infty} g(x) \ dx$$

Example

$$ \text{Evaluate } \int_{0}^{\infty} xe^{-x}\ dx$$
If the antiderivative is not obvious, perhaps compute it first.
$$ \int xe^{-x}\ dx = -x e^{-x} - \int (-e^{-x})\ dx $$
$$ = -xe^{-x} + \int e^{-x} \ dx = -xe^{-x} -e^{-x} + C$$
$$ u = x, \ du = dx$$
$$ v = - e^{-x}, \ dv = e^{-x} dx$$

$$ \int_{0}^{\infty} xe^{-x}\ dx= \lim\limits_{r\to\infty} \int_{0}^{r} (xe^{-x})\ dx = \lim\limits_{r\to\infty} \left. (-xe^{-x} - e^{-x} )\right\vert_{0}^{r} = \lim\limits_{r\to\infty}(-re^{-r} - e^{-r} + 1) $$
$$= \lim_{r\to\infty} ((-e^{-r}) (-r - 1) + 1) = \lim\limits_{r\to\infty} \left(\frac{r + 1}{e^r} \right) + 1 = \lim\limits_{r\to\infty} \left(\frac{1}{e^r}\right) + 1 = 1$$

Type II: Discontinuous Integrands

Example

Evaluate the following
$$ \int_0^3 \frac{1}{(x-1)^2}$$
Since $f(x)$ has a VA at $x=1$, this integral is improper.

$$ \int_0^3 \frac{1}{(x-1)^2} dx = \int_0^1 \frac{1}{(x-1)^2} + \int_1^3 \frac{1}{(x-1)^2}$$

$$ \int_0^1 \frac{1}{(x-1)^2} = \lim\limits_{r\to1^{-}} \left(\int_0^r \frac{1}{(x-1)^2} dx\right) = \lim\limits_{r\to1^{-}} \left. \left(\frac{-1}{x-1} \right\vert_0^r\right) = \lim\limits_{r\to1^{-}} \left(\frac{-1}{r-1} + 1\right) = + \infty$$

We found $\int_0^1 \frac{1}{(x-1)^2} dx$ diverges, so we can say the whole integral diverges.

The Comparison Test

Suppose f and g are continuous functions on $[a,\infty]$ and that $g(x) \ge f(x) \ge 0$ for all $x\ge a$

If $\int_a^{\infty} g(x) dx$ converges, then $\int_a^{\infty} f(x) dx$ also converges.
If $\int_a^{\infty} f(x) dx$ diverges, then $\int_a^{\infty} g(x) dx$ also diverges.

However, if we know $\int_a^\infty g(x)dx$ diverges, we cannot use C.T.

If we know $\int_a^\infty g(x)dx$ converges, we can’t use C.T. as well.

Example

Determine if the integral $\int_2^\infty \frac{1}{x-sin^2(x)} dx$ converges or diverges.

Use the Comparison Test.

Let the new function be the ratio of the leading terms (term with most rapid growth) of the numerator and denominator.

$$ f(x) = \frac{1}{x-\sin^2(x)} $$
$$ g(x) = \frac{1}{x}$$
since $\sin^2(x)$ is just bounded to $[0 , -1]$.

Notice then that $\frac{1}{x-\sin^2(x)} \ge \frac{1}{x} \ge 0$

$$\frac{1}{x-\sin^2(x)} \ge \frac{1}{x} \ge 0$$

$$ \int_2^\infty \frac{1}{x} dx = \lim\limits_{r\to\infty} \left(\int_2^r \frac{1}{x}\right) = \lim\limits_{r\to\infty} \left. \left(ln(|x|)\right\vert_2^r\right)$$
$$ = \lim\limits_{r\to\infty} (ln(r) - ln(2)) = \infty$$

Since $\frac{1}{x-\sin^2(x)} dx \ge \frac{1}{x} \forall x \ge 2$ and $\int_2^\infty \frac{1}{x}dx$ also diverges, we can say that $\int_2^\infty \frac{1}{x-\sin^2(x)} dx$ also diverges by the Comparison Test.

The Comparison Test also works on Type II Improper Integrals.

11.1: Sequences

Sequence

Definition: A sequence of real numbers whose domain is (some infinite number of the non negative integers)

Customary to write $a_n$ instead of $a(n)$

Can also be represented by recursion.

Limits of a Sequence

Definition: Let $\{a_n\}$ be a sequence. We say that $\lim\limits_{n\to\infty} a_n = L$ if we can make the value of $a_n$ arbitrarily close to $L$ by choosing $n$ sufficiently large.

Theorem: Let $f$ be a function from $[1, \infty]$ to $\mathbb{R}$. Define a sequence ${a_n}$ by the formula $a_n = f(n)$. If $\lim\limits_{x\to\infty} f(x) = L$, then $\lim\limits_{n\to\infty} a_n = L$.

Examples

$$ \lim\limits_{n\to\infty} \left(\frac{\ln(n)}{n^2}\right)$$
$$ \lim\limits_{n\to\infty} \left(\frac{\ln(n)}{n^2}\right)=\lim\limits_{x\to\infty} \left(\frac{\ln(x)}{x^2}\right) $$
Now you can use L’Hopital’s rule
$$ \lim\limits_{x\to\infty} \frac{1/x}{2x} = 0$$

Squeeze Theorem also applies as well as Limit Laws

Limit of Sequence DNE

If $a_n > 0$ for all n and $\lim\limits_{n\to\infty \neq 0}$, then $\lim\limits_{n\to\infty} (-1)^n a_n$ Does not exist.

Example

$$ \lim\limits_{n\to\infty} \left(\frac{(-1)^n n}{n+3}\right)$$
Limits does not exist

Increasing, Decreasing, Monotone

Let $\left. a_n \right\vert_{n=1}^{\infty}$ be a sequence

We say that the sequence is increasing if $a_n \le a_{n+1} \forall n$
We say that the sequence is decreasing if $a_n \ge a_{n+1} \forall n$
We say that the sequence is monotone if $a_n$ is increasing or decreasing for all n

Find if sequence is monotone

You can substitute $a_{n+1}$ and see if a sequence is increasing or decreasing compared to $a_n$
Find the derivative of $a_n$

Upper and Lower Bounds

If S is a subset of $\mathbb{R}$

S is bounded above if there is a number, B, so that $\forall x \in S, x \le B$. B is called an upper bound of S.
S is bounded below if there is a number, B, so that $\forall x \in S, x \ge B$. B is called a lower bound of S.

Example

$$S = [2, 9)$$
12 is an upper bound of $S$

9 is the least upper bound

Axiom of Completeness

Any nonempty subset of $\mathbb{R}$ which is bounded above has a least upper bound.
Any nonempty subset of $\mathbb{R}$ which is bounded below has a greatest upper bound.

Real numbers include rational numbers + everything else to keep the number line continuous (irrational numbers)

Monotone Convergence Theorem

Every bounded monotone sequence converges.

Proof: way too long

Limit of Factorial Sequences

Limit of Factorial over Factorial

Just express the larger factorial as the smaller one

Limit of Exponential and Factorial

$$ \lim\limits_{n\to\infty} \frac{r^n}{n!} = 0$$
for any real number $r$
Proof: Use squeeze theorem for all the cases: $r < 1$, $r > 0$, $r < 0$, etc.

$$ \lim\limits_{n\to\infty} \frac{n!}{r^n} = \infty$$

11.2: Series

Geometric Series

If $r$ is a real number and $r \neq 1$, a geometric sum is as follows
$$ \sum_{n=0}^{\infty} r^n$$

Nth partial sum of the series
$$S_n = \sum_{n=0}^{N} r^n = 1 + r + r^2 + r^3 … r^N$$

$$\sum_{n=0}^{\infty} r^n = \frac{1}{1-r} $$
if $|r| < 1$

Geometric Sum Formula

$$ \sum_{n=k}^{\infty} r^n = \frac{r^k}{1-r} $$
if $|r| < 1$

Divergence Test

If $\lim\limits_{a\to\infty} a_n \neq 0$, then $\sum_{n=1}^{\infty} a_n$ diverges.
The converse is not true.

11.3: The Integral Test

Assuming $a_n$ can be a continuous function, is positive, and decreasing:
$\sum_{n=1}^{\infty} a_n$ converges iff $\int_{1}^{\infty} f(x)dx$ also converges.

You can use the integral test for divergence as well, if $a_n$ can be a continuous, positive, and decreasing function.

The P-Test

p is a real number
$$ \sum_{n=1}^{\infty} \frac{1}{n^p}$$
converges iff $p > 1$

Proof: Use Divergence Test for $p \ge 0$ and Integral Test

11.4: Comparison Tests

Direct Comparison Test

$$ 0 \leq a_n \leq b_n$$

$$\text{If } \sum_{n=1}^{\infty} b_n \text{ converges, then } \sum_{n=1}^{\infty} a_n \text{ also converges}$$
$$\text{If } \sum_{n=1}^{\infty} a_n \text{ diverges, then } \sum_{n=1}^{\infty} b_n \text{ also diverges}$$

Proof: For convergence use the Monotone Convergence Theorem by showing $a_n$ is bounded and monotone (increasing). For divergence, if we know that $a_n \le b_n$, we also know that $\sum\limits_{n=1}^{\infty} a_n \le \sum\limits_{n=1}^{\infty} b_n$.

Limit Comparison Test

If $a_n$ and $b_n$ are positive for all n and
$$ \lim\limits_{n\to\infty} \left(\frac{a_n}{b_n}\right) = L > 0$$
and $L$ is finite.
Then $\sum_{n=1}^{\infty} a_n$ converges iff $\sum_{n=1}^{\infty} b_n$ converges

If $\sum_{n=1}^{\infty} b_n$ diverges iff $\sum_{n=1}^{\infty} a_n$ diverges.

Example

Determine if $\sum_{n=1}^{\infty} \frac{1}{n^{1+(1/n)}}$ converges or diverges.

$$ a_n = \frac{1}{n^{1+(1/n)}},\ b_n = \frac{1}{n}$$

$$ \lim\limits_{n\to\infty} \left(\frac{a_n}{b_n}\right) = \lim\limits_{n\to\infty} \left(\frac{1}{n^{1+(1/n)} \cdot \frac{n}{1}}\right) = \lim\limits_{n\to\infty} \left(\frac{1}{n^{1/n}}\right) = \lim\limits_{n\to\infty}e^{ln(n^{-1/n})}$$
$$ = e^{(\lim\limits_{n\to\infty} -ln(n)/n)}= e^{(\lim\limits_{n\to\infty} -(1/n)/n)} = e^0 = 1 $$

$$b_n = \frac{1}{n}\text{ diverges with the P-test with } p = 1$$
Since $b_n$ diverges and $\lim\limits_{n\to\infty} \left(\frac{a_n}{b_n}\right) = L > 0$, $a_n$ also diverges.

11.5: The Alternating Series Test

If the sequence $a_n$ is a decreasing sequence with non-negative terms, and $\lim\limits_{n\to\infty} a_n = 0$, then $\sum_{n=1}^{\infty} (-1)^{n+1} a_n$ converges.

The test does not tell us whether a series diverges. It is inconclusive if the above is not true.

Proof:

Prove using the Monotone Convergence Theorem
Find that the sequence is bounded and monotone for odd indexes and even indexes separately

11.6: Absolute Convergence, and the Ratio Test & Root Test

Absolute Convergence

$\sum\limits_{n=1}^{\infty} |a_n|$ converges.
Lets you rearrange the terms without messing up the sum.

Conditional Convergence

$\sum\limits_{n=1}^{\infty} a_n$ converges, but
$\sum\limits_{n=1}^{\infty} |a_n|$ diverges.

Proposition

If $\sum\limits_{n=1}^{\infty} |a_n|$ converges, then $\sum\limits_{n=1}^{\infty} a_n$ also converges.

Suggested Steps to Find if Series is Absolutely Convergent, Conditionally Convergent, or Neither

Check for Absolute Convergence (since this might also imply convergence of the regular series)
If the series is not absolutely convergent, then test for conditional convergence with the alternating series test.
Try the divergence Test

d’Alembert’s Ratio Test

If $\lim\limits_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = L$, where $a_n$ has nonzero terms

If L < 1, then $\sum_{n=1}^{\infty} a_n$ converges absolutely.
If L > 1, then $\sum_{n=1}^{\infty} a_n$ diverges.
If L = 1 or nonexistent (not infinity tho), the ratio test is inconclusive.

Use for factorials or where e to the power of n is in the denominator

Proof:
Compare $a_n$ with geometric series

Cauchy’s Root Test

If $\lim\limits_{n\to\infty}\sqrt[^n]{|a_n|} =L$ then

If $L < 1$, then $\sum\limits_{n=1}^{\infty} a_n$ converges absolutely.
If $L > 1$, then $\sum_{n=1}^{\infty} a_n$ diverges.
If $L=1$ or L is nonexistent (not infinity tho), the Root Test is inconclusive (series can be absolutely convergent, convergent, or divergent. This is what will happen if $\sum\limits_{n=1}^{\infty} a_n$ happens to be conditionally convergent.

Useful for terms with functions that vary with n in both numerator and denominator.

Proof: Similar to above.

11.7: Convergence Testing Strategy

How to find if $\sum\limits_{n=1}^{\infty} a_n$ is convergent:

If we can easily see that $\lim_{n\to\infty} a_n \neq 0$, use the Divergence Test. Otherwise continue down the list.
If $\sum\limits_{n\to\infty} a_n$ is a P-series or a geometric series, or a constant multiple of either, use the P-Test or the Geometric Series Formula.
If $a_n$ is an algebraic function of $n$ (function using entirely algebraic operations), try the Limit Comparison Test, so that $b_n$. is the ratio of leading terms from the denominator and numerator of $a_n$.
If $\sum\limits_{n=1}^{\infty}$ is similar to, but not actually a geometric series (e.g. $a_n = \frac{2^n-n^3}{5^n + 1}$), we should consider the Comparison Tests.
If $a_n = (-1)^n c_n$ or similar, where $c_n$ is non-negative, try the Alternating Series Test.
If $a_n$ contains a factorial or other product, we should consider the Ratio Test.
If $a_n = (c_n)^n$ for some sequence $c_n$ (that is, our terms are exponential form with varying base), we should try the Root Test.
Use the Integral Test if $a_n = \frac{1}{n(\ln(n))^p}$.

Helpful Notes

$\lim\limits_{n\to\infty} \left(1 + \frac{A}{n}\right)^n = e^A$
For Comparison Test
- $1 \le n \le ln(e^n + n)$ since $1\le ln(e^n) \le ln(e^n + n)$ for large enough values of $n$
- The above also works of n on the right side is replaced with any positive power of n (e.g. $\sqrt{n}$).
- $−1 ≤ \cos(n) ≤ 1$ for all n

Helpful Limit Formulas

$\lim\limits_{n\to\infty} \frac{\ln(n)}{n^c} = 0$ if $c > 0$
$\lim\limits_{n\to\infty} \sqrt[^n]{|P(n)|} = 1$ for any nonzero polynomial $P(n)$
$\lim\limits_{n\to\infty} \frac{n^c}{e^n} = 0$ for all $c \in \mathbb{R}$
$\lim\limits_{n\to\infty} \frac{b^n}{n!} = 0$ for all $b \in \mathbb{R}$
$\lim\limits_{n\to\infty} \left(\left(1+ \frac{b}{n}\right)^n\right) = e^b$ for all $b\in\mathbb{R}$

11.8: Power Series

Theorem

Let the following be a power series with a center, $c$.
$$ F(x) = \sum\limits_{n=0}^{\infty} a_n (x - c)^n$$

$F(x)$ converges only when $x=c$.
$F(x)$ converges for all $x \in \mathbb{R}$
There is a positive number $R$ such that $F(x)$ converges absolutely whenever $|x-c| < R$ but F(x) diverges whenever $|x-c| > R$

The number $R$ is the radius of convergence of the power series.

For case 3, we must manually check $F(x-R)$ and $F(c+R)$ for convergence when $|x-c| = R$ using 11.3-11.6 techniques.

To get the radius of convergence:

Theorem A

Let $F(x) = \sum\limits_{n=0}^{\infty} a_n (x-c)^n$ be a power series. Suppose that $\lim\limits_{n\to\infty}\sqrt[^n]{|a_n|} = \rho$. Then the radius of convergence of $F(x)$ is $R = \frac{1}{\rho}$

Theorem B

Let $F(x) = \sum\limits_{n=0}^{\infty} a_n (x-c)^n$ be a power series. Suppose that $\lim\limits_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| = \rho$. Then the radius of convergence of $F(x)$ is $R = \frac{1}{\rho}$

Finding Interval of Convergence

The domain of the function $F(x) = \sum\limits_{n=0}^{\infty} a_n (x - c)^n$ is the interval of convergence of a power series. To find the interval:

Make note of $c$
Find the radius of convergence of F, $R$ denotes the radius of convergence

If $R=0$, then F only converges at c, so the interval is just the set containing c: {c}
If R is infinite, then F converges everywhere and the interval is $(-\infty, \infty)$

If R is positive and finite, the interior of the radius of convergence is $(c - R, c + R)$.

Evaluate $F(c-R)$ and $F(c + R)$ and test for convergence. If one of them converges, change the bracket of the interval to a square bracket.

11.9: Representation of Functions By Power Series

Geometric Series Formula:
If $|x| < 1$ then $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$

We are given some function, $f$, which we can rewrite in the form
$$f(x) = \frac{kx^m}{1-g(x)} $$
where $k$ and $m$ are constants, and $g(x)$ is some function of $x$. $g(x)$ will generally be a constant power of x.
The power series expansion of $f$, will be correct when $|g(x)| < 1$

Example

$$ f(x) = \frac{1}{3-x} = \frac{1}{3\left(1 - \frac{1}{3}x\right)} = \frac{1}{3} \cdot \frac{1}{1 - \left(\frac{1}{3} x\right)}$$

$$ f(x) = \frac{1}{3} \sum_{n=0}^{\infty} \left(\frac{1}{3} x\right)^n = \sum\limits_{n=0}^{\infty} \left(\frac{1}{3}\right)^{n+1}(x^n) $$
$$ f(x) = \sum_{n=0}^{\infty} \frac{x^n}{3^{n+1}}$$

The above expansion is correct when $|g(x)| < 1$ so
$$\left|\frac{1}{3} x\right| < 1$$
$$-1 <\frac{1}{3} x < 1$$
$$-3 < x < 3$$

Example

$$ f(x) = \frac{3}{5 - x^2} = 3 \frac{1}{5\left(1 - \frac{1}{5}x^2\right)} = \frac{3}{5} \cdot \frac{1}{1 - \left( \frac{1}{5} x^2\right)} $$
$$ f(x) = \frac{3}{5} \sum_{n=0}^{\infty} \left(\frac{1}{5}x^2\right)^n = \frac{3}{5} \sum_{n=0}^{\infty} \left(\frac{1}{5}\right)^n x^{2n}$$

$$ f(x) = \sum_{n=0}^{\infty}\frac{3}{5^{n+1}} x^{2n}$$

Holds when the following is true:

$$ \left|\frac{1}{5} x^2\right| < 1$$
$$ |x^2| < 5$$
$$ |x|^2 < 5 \implies |x| < \sqrt{5}$$
$$ x \in (-\sqrt{5}, \sqrt{5}) $$

Theorem

Let $F(x) = \sum_{n=0}^{\infty} a_n (x -c)^n$ be a power series with radius of convergence $R \neq 0$. Then $F$ is differentiable on the interval $(c- R, c+R)$, and
$$ F’(x) = \sum_{n=1}^{\infty} na_n (x -c)^{n-1} $$
$$ \int F(x) dx = K + \sum_{n=0}^{\infty} \frac{a_n}{n+1} (x -c)^{n+1} $$
where K is a constant.

Notice in the $F’(x)$ example, the bound starts at 1, to ensure we don’t get a negative power.

The theorem basically says we can move around the differention/integration symbols and swap around the place of them with the summation sign.

Example

Find a power series formula for $f(x) = \ln(1 +x)$
$$ f’(x) = \frac{1}{1+x} = \frac{1}{1 - (-x)} = \sum_{n=0}^{\infty} (-x)^n = (-1)^n x^n$$
$$ f(x) \in \int \sum_{n=0}^{\infty} (-1)^n x^n dx = \sum_{n=0}^{\infty} \left(\int (-1)^n x^n dx\right) = K + \sum_{n=0}^{\infty} \left(\frac{(-1)^n}{n+1} x^{n+1}\right)$$

For which K will we have
$$ K + \sum_{n=0}^{\infty} \left(\frac{(-1)^n}{n+1} x^{n+1}\right) = \ln(1+x)$$
Evaluate both sides at the center of the power series.

$$ K + \sum_{n=0}^{\infty} \left(\frac{(-1)^n}{n+1} 0^{n+1}\right) = \ln(1 + 0)$$
$$ K + 0 = \ln(1 + 0)$$
$$ K = 0$$
Thus

$$ 0 + \sum_{n=0}^{\infty} \left(\frac{(-1)^n}{n+1} x^{n+1}\right) = \ln(1 + x)$$

11.10: Taylor and Maclaurin Series

If $f$ is representable by a power series on the interval $(c-R, c+R)$ with $R>0$. Let’s say that $f(x) = \sum\limits_{n=0}^{\infty} a_n(x-c)^n$ throughout the interval. Since power series are infinitely differentiable in the interval of convergence, $f$ is infinitely differentiable at $c$.

$$ f’(x) = \frac{d}{dx} \sum_{n=0}^{\infty} a_n(x-c)^n = \sum_{n=0}^{\infty} n \cdot a_n (x-c)^{n-1} \implies f^{(1)}(c) = 1\cdot a_1$$
$$ f^{''}(x) = \frac{d}{dx} \sum_{n=0}^{\infty} na_n(x-c)^{n-1} = \sum_{n=0}^{\infty} n(n-1) \cdot a_n (x-c)^{n-2} \implies f^{(2)}(c) = 2\cdot1\cdot a_2$$
$$ f^{(3)}(x) = \frac{d}{dx} \sum_{n=0}^{\infty} n(n-1)a_n(x-c)^{n-2} = \sum_{n=0}^{\infty} n(n-1)(n-2) \cdot a_n (x-c)^{n-3} \implies f^{(3)}(c) = 3\cdot 2\cdot1\cdot a_2$$

Generally,
$$ f^{(n)}(c) = n!\cdot a_n$$

Theorem: Is the Power Expansion of f Unique?

If $f$ is representable by a power series on the interval $(c-R, c+R)$ for some positive number $R$ such that $f(x) = \sum\limits_{n=0}^{\infty} a_n(x-c)^n$ throughout the interval, then

$$ a_n = \frac{f^{(n)} (c)}{n!}$$

Hence, the power series expansion of $f$ with a given center is $unique$.

Definitions of Taylor & Maclaurin

Let $f$ be a function which is infinitely differentiable at the number $c$.

The Taylor series for $f$ centered at $c$ is $T(x) = \sum\limits_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n$
The Maclaurin series for $f$ is just its Taylor series centered at 0. $M(x) = \sum\limits_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}(x)^n$
The Nth Taylor coefficient for $f$ at $c$ is $\frac{f^{(n)}(c)}{n!}$
The Nth Maclaurin coefficient for $f$ at $c$ is $\frac{f^{(n)}(0)}{n!}$

Theorem: When does a Taylor Series of a Function equal that Function?

If $c \in \mathbb{R}$ and $R > 0$ and $f$ is infinitely differentiable on $(c-R, c+R)$ and the following condition is met
$$\forall n > 0, \forall x \in (c-R, c+R) \ \exists K : |f^{(n)} (x) \le K|$$
then we can say that $f(x)$ is equal to its Taylor series centered at $c$ in that interval.

The Maclaurin expansion of any polynomial is just the function.