## Series Convergence Tests with Prototypical Examples

This is a note written for my Fall 2016 Math 100 class at Brown University. We are currently learning about various tests for determining whether series converge or diverge. In this note, we collect these tests together in a single document. We give a brief description of each test, some indicators of when each test would be good to use, and give a prototypical example for each. Note that we do justify any of these tests here — we’ve discussed that extensively in class. [But if something is unclear, send me an email or head to my office hours]. This is here to remind us of the variety of the various tests of convergence.

A copy of just the statements of the tests, put together, can be found here. A pdf copy of this whole post can be found here.

In order, we discuss the following tests:

- The $n$th term test, also called the basic divergence test
- Recognizing an alternating series
- Recognizing a geometric series
- Recognizing a telescoping series
- The Integral Test
- P-series
- Direct (or basic) comparison
- Limit comparison
- The ratio test
- The root test

## The $n$th term test

### Statement

Suppose we are looking at $\sum_{n = 1}^\infty a_n$ and

\begin{equation}

\lim_{n \to \infty} a_n \neq 0. \notag

\end{equation}

Then $\sum_{n = 1}^\infty a_n$ does not converge.

### When to use it

When applicable, the $n$th term test for divergence is usually the easiest and quickest way to confirm that a series diverges. When first considering a series, it’s a good idea to think about whether the terms go to zero or not. But remember that if the limit of the individual terms is zero, then it is necessary to think harder about whether the series converges or diverges.

### Example

Each of the series

\begin{equation}

\sum_{n = 1}^\infty \frac{n+1}{2n + 4}, \quad \sum_{n = 1}^\infty \cos n, \quad \sum_{n = 1}^\infty \sqrt{n} \notag

\end{equation}

diverges since their limits are not $0$.

## Recognizing alternating series

### Statement

Suppose $\sum_{n = 1}^\infty (-1)^n a_n$ is a series where

- $a_n \geq 0$,
- $a_n$ is decreasing, and
- $\lim_{n \to \infty} a_n = 0$.

Then $\sum_{n = 1}^\infty (-1)^n a_n$ converges.

Stated differently, if the terms are alternating sign, decreasing in absolute size, and converging to zero, then the series converges.

### When to use it

The key is in the name — if the series is alternating, then this is the goto idea of analysis. Note that if the terms of a series are alternating and decreasing, but the terms *do not* go to zero, then the series diverges by the $n$th term test.

### Example

Suppose we are looking at the series

\begin{equation}

\sum_{n = 1}^\infty \frac{(-1)^n}{\log(n+1)} = \frac{-1}{\log 2} + \frac{1}{\log 3} + \frac{-1}{\log 4} + \cdots \notag

\end{equation}

The terms are alternating.

The sizes of the terms are $\frac{1}{\log (n+1)}$, and these are decreasing.

Finally,

\begin{equation}

\lim_{n \to \infty} \frac{1}{\log(n+1)} = 0. \notag

\end{equation}

Thus the alternating series test applies and shows that this series converges.