Sequences and Series - Examples and Constructions

Convergence tests provide systematic methods for determining whether series converge without computing exact sums. These tests are essential tools for analysis and applications throughout mathematics.

TheoremIntegral Test

Let $f$ be a continuous, positive, decreasing function on $[1, \infty)$ with $f(n) = a_n$ . Then: $\sum_{n=1}^\infty a_n \text{ and } \int_1^\infty f(x)\,dx$ either both converge or both diverge.

Examplep-Series Test

The p-series $\sum_{n=1}^\infty \frac{1}{n^p}$ converges if $p > 1$ and diverges if $p \leq 1$ .

Proof: Use the integral test with $f(x) = \frac{1}{x^p}$ : $\int_1^\infty \frac{1}{x^p}\,dx = \lim_{t \to \infty} \left[\frac{x^{-p+1}}{-p+1}\right]_1^t$

If $p > 1$ : $-p+1 < 0$ , so the integral converges to $\frac{1}{p-1}$
If $p \leq 1$ : the integral diverges

Therefore $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, but $\sum_{n=1}^\infty \frac{1}{n}$ diverges.

TheoremComparison Test

Suppose $0 \leq a_n \leq b_n$ for all $n \geq N$ .

If $\sum b_n$ converges, then $\sum a_n$ converges
If $\sum a_n$ diverges, then $\sum b_n$ diverges

ExampleUsing Direct Comparison

Test $\sum_{n=1}^\infty \frac{1}{n^2 + 1}$ for convergence.

For $n \geq 1$ : $n^2 + 1 > n^2$ , so $\frac{1}{n^2 + 1} < \frac{1}{n^2}$ .

Since $\sum \frac{1}{n^2}$ converges (p-series with $p = 2 > 1$ ), by comparison, $\sum \frac{1}{n^2+1}$ converges.

TheoremLimit Comparison Test

Suppose $a_n, b_n > 0$ for all $n$ and $\lim_{n \to \infty} \frac{a_n}{b_n} = L$ where $0 < L < \infty$ .

Then $\sum a_n$ and $\sum b_n$ either both converge or both diverge.

ExampleLimit Comparison

Test $\sum_{n=1}^\infty \frac{2n^2 + 3}{n^3 + n + 1}$ for convergence.

Compare with $b_n = \frac{1}{n}$ (which diverges): $\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{2n^2 + 3}{n^3 + n + 1} \cdot n = \lim_{n \to \infty} \frac{2n^3 + 3n}{n^3 + n + 1} = 2$

Since the limit is positive and finite, and $\sum \frac{1}{n}$ diverges, $\sum a_n$ diverges.

TheoremRatio Test

Let $a_n > 0$ and $L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ .

If $L < 1$ , then $\sum a_n$ converges
If $L > 1$ (or $L = \infty$ ), then $\sum a_n$ diverges
If $L = 1$ , the test is inconclusive

ExampleRatio Test Application

Test $\sum_{n=1}^\infty \frac{2^n}{n!}$ for convergence.

$L = \lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{2^{n+1}/(n+1)!}{2^n/n!} = \lim_{n \to \infty} \frac{2 \cdot n!}{(n+1)!} = \lim_{n \to \infty} \frac{2}{n+1} = 0$

Since $L = 0 < 1$ , the series converges.

TheoremRoot Test

Let $a_n \geq 0$ and $L = \lim_{n \to \infty} \sqrt[n]{a_n}$ .

If $L < 1$ , then $\sum a_n$ converges
If $L > 1$ , then $\sum a_n$ diverges
If $L = 1$ , the test is inconclusive

TheoremAlternating Series Test (Leibniz)

If $\{b_n\}$ is a decreasing sequence with $\lim_{n \to \infty} b_n = 0$ , then $\sum_{n=1}^\infty (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + \cdots$ converges.

ExampleAlternating Harmonic Series

The series $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ converges by the Alternating Series Test:

$b_n = \frac{1}{n}$ is decreasing
$\lim_{n \to \infty} \frac{1}{n} = 0$

(The sum is $\ln 2$ , though the test doesn't give us this value.)

DefinitionAbsolute and Conditional Convergence

A series $\sum a_n$ is:

Absolutely convergent if $\sum |a_n|$ converges
Conditionally convergent if $\sum a_n$ converges but $\sum |a_n|$ diverges

Absolute convergence implies convergence, but not conversely.

ExampleConditional Convergence

$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}$ converges (Alternating Series Test) but $\sum_{n=1}^\infty \frac{1}{n}$ diverges (harmonic series).

Therefore it is conditionally convergent.

These convergence tests form a comprehensive toolkit for analyzing infinite series across mathematics and applications.