Series & Sequences

A sequence $\{a_n\}$ is an ordered list: $a_1, a_2, a_3, \ldots$ The sequence converges if $\lim_{n\to\infty} a_n = L$ for some finite $L$.

Common sequences: $a_n = 1/n \to 0$, $a_n = (1+1/n)^n \to e$, $a_n = (-1)^n$ diverges (oscillates).

Tools for evaluating sequence limits: L'Hôpital's Rule (treat $n$ as a continuous variable $x$), Squeeze Theorem, and the fact that exponentials dominate polynomials which dominate logarithms: $\ln n \ll n^p \ll a^n \ll n! \ll n^n$ for $a > 1$.

A sequence is monotonic if it is always increasing or always decreasing. A bounded monotonic sequence always converges (Monotone Convergence Theorem).

A series $\sum_{n=1}^{\infty} a_n$ is the limit of its partial sums: $S_N = \sum_{n=1}^{N} a_n$. If $\lim_{N\to\infty} S_N$ exists and is finite, the series converges.

Key distinction: a sequence $\{a_n\}$ converging to $0$ is necessary for the series to converge, but not sufficient. $a_n = 1/n \to 0$, yet $\sum 1/n$ diverges.

The harmonic series $\sum 1/n$ diverges. This is one of the most important facts in the theory of series. It shows that terms going to zero isn't enough.

A geometric series has the form $\sum_{n=0}^{\infty} ar^n$ where $a$ is the first term and $r$ is the common ratio.

It converges if and only if $|r| < 1$, and the sum is $\frac{a}{1-r}$.

Example: $\sum_{n=0}^{\infty} \frac{3}{2^n} = \frac{3}{1-1/2} = 6$.

This is the only common series where we can easily compute the exact sum. Many convergence tests compare other series to geometric ones.

A telescoping series is one where most terms cancel in the partial sum.

Example: $\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \left(\frac{1}{n} - \frac{1}{n+1}\right)$.

The partial sum is $S_N = 1 - \frac{1}{N+1}$. As $N \to \infty$, $S_N \to 1$.

To recognize telescoping: use partial fractions to decompose the general term. If consecutive terms cancel, you have a telescoping series.

If $\lim_{n\to\infty} a_n \neq 0$, then $\sum a_n$ diverges. Period.

This should always be your first check. It's quick and catches obvious divergence.

If $\lim_{n\to\infty} a_n = 0$, the test is inconclusive. You need another test. (Remember: $\sum 1/n$ has $a_n \to 0$ but diverges.)

A p-series is $\sum_{n=1}^{\infty} \frac{1}{n^p}$.

Converges if $p > 1$ and diverges if $p \leq 1$.

$p=1$: harmonic series $\sum 1/n$ (diverges). $p=2$: $\sum 1/n^2 = \pi^2/6$ (converges). $p=1/2$: $\sum 1/\sqrt{n}$ (diverges).

p-series are the benchmark for comparison tests. When you see $1/n^p$-like terms, think p-series.

If $f(x)$ is positive, continuous, and decreasing for $x \geq N$, and $a_n = f(n)$, then $\sum_{n=N}^{\infty} a_n$ and $\int_N^{\infty} f(x)\,dx$ either both converge or both diverge.

The integral test does not give the sum of the series, only whether it converges.

Example: test $\sum 1/n^2$. Compare with $\int_1^{\infty} 1/x^2\,dx = 1$. The integral converges, so the series converges.

This is how we prove the p-series convergence criteria: $\int_1^{\infty} x^{-p}\,dx$ converges iff $p > 1$.

Direct comparison: if $0 \leq a_n \leq b_n$ for all $n$ and $\sum b_n$ converges, then $\sum a_n$ converges. If $\sum a_n$ diverges, then $\sum b_n$ diverges.

Limit comparison: if $\lim_{n\to\infty} a_n / b_n = L$ where $0 < L < \infty$, then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

Strategy: compare your series to a known p-series or geometric series. Pick $b_n$ by keeping only the dominant terms.

Example: test $\sum \frac{n}{n^3+1}$. For large $n$, this behaves like $\frac{n}{n^3} = \frac{1}{n^2}$. Limit comparison with $\sum 1/n^2$ (converges) confirms convergence.

Ratio test: compute $L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|$. If $L < 1$: converges absolutely. $L > 1$: diverges. $L = 1$: inconclusive.

Root test: compute $L = \lim_{n\to\infty} \sqrt[n]{|a_n|}$. Same criteria as the ratio test.

The ratio test excels for series with factorials (like $n!/3^n$) or products. The root test is ideal for terms raised to the $n$th power (like $(2/3)^n$).

Warning: both tests are inconclusive when $L=1$. This happens for all p-series, so you can't use ratio/root tests on $\sum 1/n^p$.

An alternating series has the form $\sum (-1)^n a_n$ where $a_n > 0$.

It converges if: (1) $a_n$ is eventually decreasing and (2) $\lim_{n\to\infty} a_n = 0$.

Example: $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - 1/2 + 1/3 - 1/4 + \cdots = \ln 2$. (The alternating harmonic series.)

Alternating series estimation: the error after summing $N$ terms is at most $|a_{N+1}|$. This gives a built-in error bound.

A series $\sum a_n$ converges absolutely if $\sum |a_n|$ converges. It converges conditionally if $\sum a_n$ converges but $\sum |a_n|$ diverges.

Absolute convergence implies convergence. So if the absolute value series converges, you're done.

Example: $\sum (-1)^n/n^2$ converges absolutely because $\sum 1/n^2$ converges.

Example: $\sum (-1)^n/n$ converges conditionally. It converges by the alternating series test, but $\sum 1/n$ diverges.

Why it matters: absolutely convergent series can be rearranged freely without changing the sum. Conditionally convergent series cannot (Riemann rearrangement theorem).

A power series centered at $a$ is $\sum_{n=0}^{\infty} c_n(x-a)^n$. It converges on some interval centered at $a$.

The radius of convergence $R$ determines where it converges: $|x-a| < R$ (converges), $|x-a| > R$ (diverges). At $|x-a| = R$ (the endpoints), you must check separately.

Find $R$ using the ratio test: $R = \lim_{n\to\infty} \left|\frac{c_n}{c_{n+1}}\right|$ (or the root test).

Inside the radius, power series behave like polynomials: you can differentiate and integrate them term by term.

Example: the geometric series $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| < 1$ is a power series with $R = 1$.

The Taylor series of $f(x)$ centered at $a$ is: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$.

A Maclaurin series is a Taylor series centered at $a=0$.

Essential Maclaurin series to memorize:

$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$ for all $x$.

$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$ for all $x$.

$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$ for all $x$.

$\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$ for $-1 < x \leq 1$.

$\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| < 1$.

The $n$th degree Taylor polynomial $T_n(x)$ approximates $f(x)$. The error is $R_n(x) = f(x) - T_n(x)$.

Taylor's inequality: $|R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}$ where $M$ is an upper bound for $|f^{(n+1)}(t)|$ on the interval between $a$ and $x$.

For alternating series, the error after $n$ terms is bounded by the first omitted term: $|R_n| \leq |a_{n+1}|$.

This is critical for applications: you can determine how many terms you need for a desired accuracy.

Substitution: replace $x$ in a known series. E.g., $e^{-x^2} = \sum \frac{(-x^2)^n}{n!} = \sum \frac{(-1)^n x^{2n}}{n!}$.

Differentiation: $\frac{d}{dx}\left[\frac{1}{1-x}\right] = \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} nx^{n-1}$.

Integration: $\int_0^x \frac{1}{1+t^2}\,dt = \arctan x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1}$.

These techniques let you derive new series without computing derivatives from scratch.

1. Divergence test first: if $a_n \not\to 0$, done (diverges).

2. Is it geometric? Check if $a_n = ar^n$.

3. Is it a p-series? Check if $a_n = 1/n^p$.

4. Does it telescope? Try partial fractions.

5. Are there factorials or $n$th powers? Try ratio or root test.

6. Does it look like a known series? Try comparison or limit comparison.

7. Is it alternating? Try the alternating series test.

8. Is $f(n) = a_n$ easy to integrate? Try the integral test.