Series Convergence Cheat Sheet

Your guide to conquering infinite series!

Convergence Tests Summary

Test Conditions & Limit Conclusion
n-th Term Test for Divergence \(\lim_{n \to \infty} a_n \neq 0\) Series Diverges. (Inconclusive if limit is 0).
p-Series
\(\sum \frac{1}{n^p}\)		 Check the value of \(p\). Converges if \(p > 1\).
Diverges if \(p \le 1\).
Alternating Series Test
\(\sum (-1)^n b_n\)		 1. \(b_{n+1} \le b_n\) (decreasing)
2. \(\lim_{n \to \infty} b_n = 0\)		 Series Converges. (Check for absolute convergence separately).
Ratio Test \(L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\) Abs. Convergent if \(L < 1\).
Divergent if \(L > 1\).
Inconclusive if \(L = 1\).
Root Test \(L = \lim_{n \to \infty} \sqrt[n]{|a_n|}\) Abs. Convergent if \(L < 1\).
Divergent if \(L > 1\).
Inconclusive if \(L = 1\).

The Ratio Test

This is a powerful test, especially for series involving factorials ($n!$) or exponentials ($k^n$).

For a series \(\sum a_n\), calculate the limit:

$$ L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| $$
  • If \(L < 1\), the series is absolutely convergent (and therefore convergent).
  • If \(L > 1\) or \(L = \infty\), the series is divergent.
  • If \(L = 1\), the Ratio Test is inconclusive. You must use a different test!

Example 1: A Convergent Series

Test the series \(\sum_{n=1}^{\infty} \frac{2^n}{n!}\) for absolute convergence.

Here, \(a_n = \frac{2^n}{n!}\) and \(a_{n+1} = \frac{2^{n+1}}{(n+1)!}\).

$$ L = \lim_{n \to \infty} \left| \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} \right| $$ $$ = \lim_{n \to \infty} \left| \frac{2 \cdot 2^n}{(n+1) \cdot n!} \cdot \frac{n!}{2^n} \right| = \lim_{n \to \infty} \left| \frac{2}{n+1} \right| = 0 $$

Since \(L = 0 < 1\), the series is absolutely convergent.

Example 2: A Divergent Series

Test the series \(\sum_{n=1}^{\infty} \frac{(-3)^n}{n^3}\).

$$ L = \lim_{n \to \infty} \left| \frac{(-3)^{n+1}}{(n+1)^3} \cdot \frac{n^3}{(-3)^n} \right| $$ $$ = \lim_{n \to \infty} \left| \frac{-3 \cdot n^3}{(n+1)^3} \right| = 3 \lim_{n \to \infty} \left( \frac{n}{n+1} \right)^3 = 3(1)^3 = 3 $$

Since \(L = 3 > 1\), the series diverges.

Example 3: An Inconclusive Result

Test the series \(\sum_{n=1}^{\infty} \frac{3n}{n+2}\).

$$ L = \lim_{n \to \infty} \left| \frac{3(n+1)}{(n+1)+2} \cdot \frac{n+2}{3n} \right| = \lim_{n \to \infty} \left| \frac{n+1}{n+3} \cdot \frac{n+2}{n} \right| = 1 $$

Since \(L = 1\), the Ratio Test is inconclusive.

What now? Try another test! The Divergence Test is a good choice here. $$ \lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{3n}{n+2} = 3 $$ Since the limit is not 0, the series diverges by the Divergence Test.

Other Key Convergence Tests

The n-th Term Test for Divergence

This should be the FIRST test you perform on any series. It can only prove divergence.

$$ \text{If } \lim_{n \to \infty} a_n \neq 0, \text{ the series } \sum a_n \text{ diverges.} $$

If the limit IS 0, this test is inconclusive. The series might converge or diverge.

p-Series Test

A quick test for series of the form \(\sum \frac{1}{n^p}\).

$$ \text{The series } \sum_{n=1}^{\infty} \frac{1}{n^p} \text{ is:} $$
  • Convergent if \(p > 1\).
  • Divergent if \(p \le 1\).

Example: \(\sum \frac{1}{n^2}\) converges (p=2), but \(\sum \frac{1}{\sqrt{n}}\) diverges (p=1/2).

Alternating Series Test (AST)

Use for series with alternating signs, like \(\sum (-1)^n b_n\).

The series converges if it meets BOTH conditions:

  1. \(b_{n+1} \le b_n\) for all n (the terms are decreasing).
  2. \(\lim_{n \to \infty} b_n = 0\).

AST only tells you about convergence, not *absolute* convergence. You must test \(\sum |a_n|\) separately to check for that.

The Root Test

A cousin of the Ratio Test. Very useful if the series involves terms raised to the n-th power.

$$ L = \lim_{n \to \infty} \sqrt[n]{|a_n|} $$
  • If \(L < 1\), the series is absolutely convergent.
  • If \(L > 1\) or \(L = \infty\), the series is divergent.
  • If \(L = 1\), the Root Test is inconclusive.