Calc III Study Set: Sequences and Series

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Sequence

A function whose domain is the set of positive integers. For example:
{5, 9, 13, 17, …} is an example of an Arithmetic Sequence (constant common difference between terms) with the formula a_n = 4n + 1 (nth term).
{1/2, 1/4, 1/8, 1/16, …} is an example of a Geometric Sequence (constant common ratio between terms) with the formula a_n = (1/2)^n (nth term).

Series

The sum of the terms in a sequence.
Examples:
Finite Series: $\sum_{n=1}^{3} \left( \frac{1}{5^n} \right) = \frac{1}{5} + \frac{1}{5^2} + \frac{1}{5^3} = \frac{31}{125}$
Infinite Series: $\sum_{n=1}^{\infty} (2n - 3) = -1 + 1 + 3 + 5 + \ldots$

Partial Sums

Partial Sums - The partial sums of a series form a sequence of real numbers defined as:
$S_1 = a_1$ $S_2 = a_1 + a_2$ $S_3 = a_1 + a_2 + a_3$ $\vdots$ $S_n = a_1 + a_2 + \cdots + a_n$
If the sequence of partial sums ${S_1, S_2, S_3, \ldots}$ has a limit $S$ as $n \to \infty$, then the series converges to $S$ (otherwise it diverges):
$\lim_{n \to \infty} S_n = S$

Limit of nth Term of a Convergent Series

Divergence Test - If the infinite series $\sum_{n=1}^{\infty} a_n$ converges, then $\lim_{n \to \infty} a_n = 0$
Note: The converse is not always true; $\lim_{n \to \infty} a_n = 0$ does not imply that $\sum_{n=1}^{\infty} a_n$ converges.

Nth Term Test for Divergence

If $\lim_{n \to \infty} a_n \neq 0$, then the infinite series $\sum_{n=1}^{\infty} a_n$ will diverge.

Convergence of Geometric Series

Convergence of Geometric Series - A geometric series can be written as:
$\sum_{n=1}^{\infty} a \cdot r^{n-1} \quad \text{or} \quad \sum_{n=0}^{\infty} a \cdot r^{n}$
All geometric series will converge to $\frac{a_1}{1-r}$ if $0 < |r| < 1$ and will diverge if $|r| \geq 1$.

Tail of a Series (Term not needed to know but need to know definition)

If a series converges, then removing a finite number of terms from the beginning of the series will not affect its convergence.

The Integral Test

If a sequence is decreasing and positive and can be represented by $a_n = f(n)$ and $f$ is continuous for $x \geq 1$, then $\sum_{n=1}^{\infty}(a_n)$ and $\int_1^{\infty} f(x)dx$ either both converge or both diverge.

P-Series

A p-series can be written as: $\sum_{n=1}^{\infty} \frac{1}{n^p} \quad \text{or} \quad \sum_{k=1}^{\infty} \frac{1}{k^p}$ This infinite series converges if (and only if) $p > 1$. Otherwise, it diverges.

Direct Comparison Test

Let $0 < a_n \leq b_n$ for all $n$. If $\sum_{n=1}^{\infty}(b_n)$ converges, then $\sum_{n=1}^{\infty}(a_n)$ converges. 2. If $\sum_{n=1}^{\infty}(a_n)$ diverges, then $\sum_{n=1}^{\infty}(b_n)$ diverges.

Direct Comparison Test (Key Idea)

When using the Direct Comparison Test, find a "similar but simpler" series to compare to. If the "larger" series converges, then the "smaller" series must also converge. If the "smaller" series diverges, then the "larger" series must also diverge. Note: if your original series is larger than a convergent series, or smaller than a divergent series, the test tells you nothing and you must use a different test.

Limit Comparison Test

If $a_n$ and $b_n$ are both positive and $\lim_{n \to \infty} \left(\frac{a_n}{b_n}\right)$ is finite and positive, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge. The trick to choosing a good $b_n$ to compare to is to take the highest power of $n$ in both the numerator and denominator and divide the numerator term by the denominator term.

Telescoping Series

A series where consecutive terms cancel each other out, leaving only the first and last terms. To solve, use partial fractions to rewrite the series, write out the first few terms to see the cancellation pattern, and the sum equals $S = \lim_{n \to \infty} S_n$ where most terms cancel. For example, $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ can be rewritten as $\sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$, where all middle terms cancel, leaving $S = 1$.

Alternating Series

An alternating series is a series where the signs alternate, written as $\sum_{n=1}^{\infty} (-1)^{n+1}(a_n) = a_1 - a_2 + a_3 - a_4 + ...$ where $a_n > 0$ for all $n$.

Alternating Series Test

An alternating series converges if BOTH conditions are met:
1. The terms are decreasing: $0 < a_{n+1} \leq a_n$ for all $n \geq 1$
2. The terms approach zero: $\lim_{n \to \infty} a_n = 0$
Then the alternating series:
$\sum_{n=1}^{\infty} (-1)^{n+1}(a_n) \quad \text{or} \quad \sum_{n=1}^{\infty} (-1)^{n}(a_n)$
converges.

Alternating Series Remainder

For a convergent alternating series, the remainder (error) between the actual sum $S$ and the partial sum $S_n$ is bounded by the first neglected term: $|R_n| = |S - S_n| \leq a_{n+1}$

Absolute Convergence

If $\sum_{n=1}^{\infty} |a_n|$ converges, then $\sum_{n=1}^{\infty} a_n$ converges. A series $\sum_{n=1}^{\infty} a_n$ converges absolutely if $\sum_{n=1}^{\infty} |a_n|$ converges. The converse is not always true — a series can converge without converging absolutely.

Absolute vs. Conditional Convergence

$\sum_{n=1}^{\infty} a_n$ is absolutely convergent if $\sum_{n=1}^{\infty} |a_n|$ converges. $\sum_{n=1}^{\infty} a_n$ is conditionally convergent if $\sum_{n=1}^{\infty} a_n$ converges, but $\sum_{n=1}^{\infty} |a_n|$ diverges.

Ratio Test

Suppose $\sum_{n=1}^{\infty} a_n$ with $a_n \neq 0$. Let $\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L$
If $L < 1$, the series converges absolutely.
If $L > 1$, the series diverges.
If $L = 1$, no conclusion can be drawn.

$\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^n = \text{ ?}$

$\lim_{n\to\infty}\left(\frac{n+1}{n}\right)^n = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e$

Maclaurin Polynomial

The $n^{\text{th}}$ Maclaurin Polynomial for $f$, centered at 0, is defined as:
$P_n(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \frac{f^{(4)}(0)x^4}{4!} + \cdots + \frac{f^{(n)}(0)x^n}{n!}$

Taylor Polynomial

The $n^{\text{th}}$ Taylor Polynomial for $f$ at $c$ is defined as:
$P_n(x) = T_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)(x-c)^2}{2!} + \cdots + \frac{f^{(n)}(c)(x-c)^n}{n!}$

Remainder of the Taylor Polynomial

With every polynomial approximation (Maclaurin or Taylor), there exists some error between the actual function value and our approximation as you move away from the center.
The actual function can be written as:
$f(x) = P_n(x) + E_n(x)$
where $E_n(x)$ is the remainder (error), defined as:
$|E_n(x)| = |f(x) - P_n(x)|$

Taylor's Theorem

If a function $f$ is differentiable through order $(n+1)$ on an interval containing $c$, then for each $x$ in the interval, there exists a $z$ between $x$ and $c$ such that:
$f(x) = f(c) + f'(c)(x-c) + \frac{f''(c)(x-c)^2}{2!} + \cdots + \frac{f^{(n)}(c)(x-c)^n}{n!} + E_n(x)$
where $E_n(x)$ is the remainder in Lagrange Form:
$E_n(x) = \frac{f^{(n+1)}(z)(x-c)^{n+1}}{(n+1)!}$
Similar to the Alternating Series Remainder, if we know properties of the $(n+1)^{\text{st}}$ derivative of $f$, we can bound the error by maximizing that $(n+1)^{\text{st}}$ derivative:
$|E_n(x)| \leq \frac{M|x-c|^{n+1}}{(n+1)!}$
where $M$ is the maximum value of the $(n+1)^{\text{st}}$ derivative between $x$ and $c$.

Taylor Series

If a function $f$ has derivatives of all orders at $x = c$, then:
$\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!}(x-c)^n = f(c) + f'(c)(x-c) + \cdots + \frac{f^{(n)}(c)}{n!}(x-c)^n + \cdots$
is called the Taylor Series for $f(x)$ at $c$. If $c = 0$, then the series is the Maclaurin Series for $f$.

Power Series

Any expression of the form:
$\sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n + \cdots$
is a power series centered at zero.
Any expression of the form:
$\sum_{n=0}^{\infty} a_n (x-c)^n = a_0 + a_1(x-c) + \cdots + a_n(x-c)^n + \cdots$
is a power series centered at $c$ (where $c$ is constant).

Convergence of Power Series

For a power series centered at $c$, one of the following is true:
1. The series converges only at $c$. [Its radius of convergence is 0.]
2. There exists an $R > 0$ such that the series converges absolutely for $|x - c| < R$ and diverges for $|x - c| > R$. [The interval of convergence is $(c - R, c + R)$.]
3. The series converges absolutely for all $x$. [The interval of convergence is $(-\infty, \infty)$.]
The Ratio Test is useful for determining the interval of convergence. Instead of:
$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$
we now use:
$\lim_{n \to \infty} \left|\frac{u_{n+1}}{u_n}\right|$
where $u$ is a function of $x$.

Geometric Power Series

A geometric power series represents a function within its interval of convergence using the geometric series formula. Since $\sum_{n=0}^{\infty} r^n = \frac{1}{1-r}$ for $|r| < 1$, we can rewrite functions in this form. For example:
$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots \quad |x| < 1$

Operations with Power Series

If $f(x) = \sum_{n=0}^{\infty} a_n x^n$ and $g(x) = \sum_{n=0}^{\infty} b_n x^n$, then the following operations are defined:
Addition/Subtraction: $f(x) \pm g(x) = \sum_{n=0}^{\infty} (a_n \pm b_n)x^n$
Multiplication by a constant: $k \cdot f(x) = \sum_{n=0}^{\infty} k a_n x^n$
Substitution: Replace $x$ with an expression, e.g. $f(x^2) = \sum_{n=0}^{\infty} a_n x^{2n}$, but the interval of convergence must be re-evaluated after substitution. For example, if $|x| < 1$, then for $f(x^2)$: $|x^2| < 1 \implies |x| < 1$ (the same here, but this will not always be the case).

Differentiation and Integration of Power Series

If $f(x) = \sum_{n=0}^{\infty} a_n (x-c)^n$, then within the interval of convergence:
Differentiation: $f'(x) = \sum_{n=1}^{\infty} n \cdot a_n (x-c)^{n-1}$
Integration: $\int f(x)\,dx = \sum_{n=0}^{\infty} \frac{a_n (x-c)^{n+1}}{n+1} + C$
Note: Differentiation and integration do not change the radius of convergence, but may affect convergence at the endpoints.

Radius of Convergence

The radius of convergence $R$ of a power series centered at $c$ is the value such that the series converges absolutely for $|x - c| < R$ and diverges for $|x - c| > R$. It is found using the Ratio Test:
$\lim_{n \to \infty} \left|\frac{u_{n+1}}{u_n}\right| < 1$
Solving this inequality for $|x - c|$ gives the radius $R$. Special cases:
- If $R = 0$, the series converges only at $x = c$.
- If $R = \infty$, the series converges for all $x$.

Interval of Convergence

The interval of convergence is the set of all $x$ values for which a power series converges. To find it:
1. Use the Ratio Test to find the radius of convergence $R$, giving the open interval $(c - R, c + R)$.
2. Check each endpoint separately by substituting into the original series and applying an appropriate convergence test (e.g. Alternating Series Test, p-series, Divergence Test).
3. Include an endpoint in the interval if the series converges there.
The final interval of convergence can be open, half-open, or closed at the endpoints.

Lagrange Error Bound

The Lagrange Error Bound gives an upper bound on the error between a function and its Taylor/Maclaurin polynomial approximation:
$|E_n(x)| \leq \frac{M|x-c|^{n+1}}{(n+1)!}$
where $M$ is the maximum value of $|f^{(n+1)}(z)|$ for all $z$ between $x$ and $c$.
To find $M$:
1. Find the $(n+1)^{\text{th}}$ derivative of $f$.
2. Determine the interval of $z$ values (all values between $x$ and $c$).
3. Maximize $|f^{(n+1)}(z)|$ on that interval. If the exact maximum is difficult to find, use a convenient upper bound instead.
4. That maximum (or upper bound) is $M$.
Example: Approximate $e^{0.1}$ using a degree 3 Maclaurin polynomial and bound the error.
- $f(x) = e^x$, so $f^{(4)}(z) = e^z$
- $z$ is between 0 and 0.1, so $e^z \leq e^{0.1} < 2$ is a safe upper bound, giving $M = 2$
$|E_3(0.1)| \leq \frac{2(0.1)^4}{4!} = \frac{0.0002}{24} \approx 0.0000083$

What’s On The Test

5 MCQs
2 FRQ
Telescoping Series
3 different series and you have to say which converges
nth term test… etc
1 Bonus (Something we did in class)

Taylor Series to Memorize:
$\sin(x)$
$\cos(x)$
$e^{x}$
$\frac{1}{1-x}$

Taylor Series to Memorize
$\sin(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \quad (-\infty, \infty)$
$\cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \quad (-\infty, \infty)$
$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \quad (-\infty, \infty)$
$$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + \cdots \quad (-1, 1