# Series Convergence Tests: Which Test to Use and When

> Every series convergence test with examples: nth-term, integral, p-series, comparison, ratio, root and alternating, plus a which-test table.

Source: https://practicecalculus.com/blog/convergence-tests
Updated: 2026-10-22

## Key takeaways

- Always try the nth-term test first: if the terms do not approach 0, the series diverges, and you are done.
- Compare to a geometric series or a p-series $\sum \frac{1}{n^p}$ (which converges exactly when $p > 1$) whenever the terms look like a power of $n$.
- Use the ratio test for factorials and exponentials, the root test for terms raised to the $n$th power, and the alternating series test for signs that flip.
- If the ratio or root test gives a limit of exactly 1, it tells you nothing. Switch to a different test.

You are staring at $\sum \frac{3^n}{n!}$ and you know there are eight different convergence tests, but which one? Most series questions are not hard once you pick the right test. The hard part is the picking. This guide covers every standard **convergence test** with a worked example, then gives you a decision table for choosing quickly.

## What does it mean for a series to converge?

A series $\sum a_n$ **converges** if its partial sums $S_N = a_1 + a_2 + \cdots + a_N$ approach a finite number as $N \to \infty$. Otherwise it **diverges**.

Most of the time you cannot compute the sum exactly ([geometric series](/blog/geometric-series) are the big exception). Convergence tests answer a simpler question: does the sum exist at all? Each test below gives a **converges**, **diverges**, or **inconclusive** verdict.

## 1. The nth-term test for divergence

$$\text{If } \lim_{n\to\infty} a_n \ne 0 \text{ (or the limit does not exist), then } \sum a_n \text{ diverges.}$$

**Example.** $\sum \frac{n}{2n+1}$. The terms approach $\frac12$, not 0. Adding infinitely many numbers close to $\frac12$ cannot give a finite total, so the series diverges.

> **Tip:** this test can **never** prove convergence. If $a_n \to 0$, the test says nothing and you must try another one.

## 2. Geometric series

$$\sum_{n=0}^\infty ar^n \text{ converges to } \frac{a}{1-r} \text{ if } |r| < 1, \text{ and diverges if } |r| \ge 1.$$

**Example.** $\sum \frac{5}{4^n}$ has $r = \frac14$, so it converges.

## 3. The p-series test

$$\sum_{n=1}^\infty \frac{1}{n^p} \text{ converges if } p > 1 \text{ and diverges if } p \le 1.$$

**Examples.** $\sum \frac{1}{n^{3/2}}$ converges ($p = 1.5$). $\sum \frac{1}{\sqrt n}$ diverges ($p = 0.5$). The **harmonic series** $\sum \frac1n$ ($p = 1$) diverges, even though its terms shrink to 0.

The harmonic series diverges extremely slowly. Its partial sums grow like $\ln N$: you need more than 12,000 terms just to pass 10. The graph shows the partial sums trapped between two logarithm curves, both of which go to infinity.

```graph
title: 'Partial sums of the harmonic series grow like ln n'
caption: 'The partial sums $H_n = 1 + \frac12 + \cdots + \frac1n$ (dots) lie between $\ln(n+1)$ and $1 + \ln n$. Both bounds grow without limit, so the harmonic series diverges.'
x: [0, 12]
y: [0, 4]
functions:
  - expr: 1 + ln(x)
    label: y = 1 + ln x
    dashed: true
  - expr: ln(x+1)
    label: y = ln(x + 1)
points:
  - [1, 1, '1']
  - [2, 1.5, '1.5']
  - [3, 1.8333]
  - [4, 2.0833]
  - [6, 2.45, '2.45']
  - [10, 2.929, '2.93']
```

## 4. The integral test

If $f$ is **positive, continuous and decreasing** for $x \ge N$, and $a_n = f(n)$, then

$$\sum_{n=N}^\infty a_n \text{ and } \int_N^\infty f(x)\,dx \text{ either both converge or both diverge.}$$

**Example.** Does $\sum_{n=2}^\infty \frac{1}{n\ln n}$ converge?

**Step 1. Check the conditions.** $f(x) = \frac{1}{x\ln x}$ is positive, continuous and decreasing for $x \ge 2$.

**Step 2. Evaluate the [improper integral](/blog/improper-integrals).** With $u = \ln x$, $du = \frac1x\,dx$:

$$\int_2^\infty \frac{dx}{x\ln x} = \lim_{b\to\infty}\Big[\ln(\ln x)\Big]_2^b = \lim_{b\to\infty}\left(\ln(\ln b) - \ln(\ln 2)\right) = \infty$$

**Step 3. Conclude.** The integral diverges, so the series diverges.

The integral test is also where the p-series rule comes from: $\int_1^\infty \frac{dx}{x^p}$ converges exactly when $p > 1$.

## 5. The direct comparison test

Suppose $0 \le a_n \le b_n$ for all large $n$.

- If $\sum b_n$ converges, then the smaller $\sum a_n$ converges.
- If $\sum a_n$ diverges, then the larger $\sum b_n$ diverges.

**Example (convergence).** $\sum \frac{1}{n^2 + 5}$. Since $n^2 + 5 > n^2$, we have $\frac{1}{n^2+5} < \frac{1}{n^2}$. The p-series $\sum \frac{1}{n^2}$ converges, so this series converges.

**Example (divergence).** $\sum_{n=3}^\infty \frac{\ln n}{n}$. For $n \ge 3$, $\ln n > 1$, so $\frac{\ln n}{n} > \frac1n$. The harmonic series diverges, so this series diverges.

## 6. The limit comparison test

Direct comparison needs an inequality in the right direction, which can be awkward. The limit comparison test only needs the terms to behave alike. If $a_n, b_n > 0$ and

$$\lim_{n\to\infty} \frac{a_n}{b_n} = L \quad \text{with } 0 < L < \infty,$$

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

**Example.** Does $\sum \frac{2n+3}{n^3+1}$ converge?

**Step 1. Pick $b_n$ from the dominant terms.** For large $n$, $\frac{2n+3}{n^3+1}$ behaves like $\frac{2n}{n^3} = \frac{2}{n^2}$. Take $b_n = \frac1{n^2}$.

**Step 2. Compute the limit.**

$$\lim_{n\to\infty}\frac{(2n+3)/(n^3+1)}{1/n^2} = \lim_{n\to\infty}\frac{2n^3 + 3n^2}{n^3 + 1} = 2$$

**Step 3. Conclude.** $L = 2$ is positive and finite, and $\sum\frac1{n^2}$ converges, so the original series converges.

## 7. The ratio test

Let

$$L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.$$

If $L < 1$ the series converges (absolutely). If $L > 1$ it diverges. If $L = 1$ the test is **inconclusive**.

**Example.** Does $\sum \frac{3^n}{n!}$ converge?

**Step 1. Form the ratio.**

$$\frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{(n+1)!}\cdot\frac{n!}{3^n} = \frac{3}{n+1}$$

**Step 2. Take the limit.** $\frac{3}{n+1} \to 0$.

**Step 3. Conclude.** $L = 0 < 1$, so the series converges.

The ratio test shines with **factorials** and **exponentials**, because most of the expression cancels. It is useless for p-series and rational functions of $n$: both $\sum\frac1n$ and $\sum\frac1{n^2}$ give $L = 1$.

## 8. The root test

Let

$$L = \lim_{n\to\infty}\sqrt[n]{|a_n|}.$$

The verdicts are the same as the ratio test: $L < 1$ converges, $L > 1$ diverges, $L = 1$ is inconclusive.

**Example.** Does $\sum \left(\frac{n}{2n+1}\right)^n$ converge?

$$\sqrt[n]{\left(\frac{n}{2n+1}\right)^n} = \frac{n}{2n+1} \to \frac12$$

Since $L = \frac12 < 1$, the series converges. Use the root test whenever the whole term is raised to the $n$th power.

## 9. The alternating series test

An alternating series has the form $\sum (-1)^{n+1} b_n$ with $b_n > 0$. It converges if

1. $b_{n+1} \le b_n$ for all large $n$ (the sizes eventually decrease), and
2. $\lim_{n\to\infty} b_n = 0$.

**Example.** The alternating harmonic series $1 - \frac12 + \frac13 - \frac14 + \cdots$ has $b_n = \frac1n$, which decreases to 0. So it converges (in fact to $\ln 2$).

**Error bound.** For a series that passes this test, stopping after $N$ terms gives an error no larger than the next term: $|S - S_N| \le b_{N+1}$. For example, the first 99 terms of the alternating harmonic series are within $\frac{1}{100}$ of $\ln 2$.

### Absolute vs conditional convergence

If $\sum |a_n|$ converges, then $\sum a_n$ converges too. We call this **absolute convergence**. The alternating harmonic series converges, but $\sum \left|\frac{(-1)^{n+1}}{n}\right| = \sum\frac1n$ diverges, so it converges only **conditionally**. By contrast, $\sum \frac{(-1)^{n+1}}{n^2}$ converges absolutely, because $\sum\frac{1}{n^2}$ converges.

## Which convergence test should you use?

Run through this table from the top. The first row that matches is usually the fastest route.

| If the series looks like...                                 | Try this test               | Example                              |
| ----------------------------------------------------------- | --------------------------- | ------------------------------------ |
| Terms that clearly do not go to 0                           | nth-term test (diverges)    | $\sum \frac{n}{2n+1}$                |
| $ar^n$, a constant ratio                                    | Geometric series            | $\sum \frac{5}{4^n}$                 |
| $\frac{1}{n^p}$                                             | p-series                    | $\sum \frac{1}{n^{3/2}}$             |
| A rational or root expression in $n$                        | Limit comparison with $\frac{1}{n^p}$ | $\sum \frac{2n+3}{n^3+1}$  |
| Close to a known series, with an easy inequality            | Direct comparison           | $\sum \frac{1}{n^2+5}$               |
| Factorials or $c^n$                                         | Ratio test                  | $\sum \frac{3^n}{n!}$                |
| The whole term raised to the $n$th power                    | Root test                   | $\sum \left(\frac{n}{2n+1}\right)^n$ |
| Signs that alternate                                        | Alternating series test     | $\sum \frac{(-1)^{n+1}}{n}$          |
| A function you can integrate easily, like $\frac{1}{n\ln n}$ | Integral test              | $\sum \frac{1}{n\ln n}$              |

> **Tip:** to find the right p-series for a comparison, keep only the highest power of $n$ in the numerator and in the denominator. $\frac{2n+3}{n^3+1}$ acts like $\frac{n}{n^3} = \frac{1}{n^2}$.

## Common convergence test mistakes

1. **Using the nth-term test to prove convergence.** $a_n \to 0$ proves nothing. The harmonic series is the classic counterexample.
2. **Treating $L = 1$ as a verdict.** For the ratio and root tests, $L = 1$ means switch tests.
3. **Comparing in the wrong direction.** Being smaller than a divergent series, or larger than a convergent one, tells you nothing.
4. **Forgetting the integral test conditions.** The function must be positive and decreasing (at least eventually).
5. **Confusing the series with the sequence.** "The terms converge to 0" and "the series converges" are different statements.
6. **Ignoring absolute values.** The ratio and root tests use $|a_n|$, and an alternating series can converge without converging absolutely.

## Practice choosing the right test

Recognizing which test fits is a pattern-matching skill, and it only gets fast with repetition. After a few dozen series you will glance at $\frac{3^n}{n!}$ and think "ratio test" before you finish reading it. The interactive lesson below mixes series of every type and gives instant feedback on both your choice of test and your conclusion.

## Frequently asked questions

### How do you know which convergence test to use?

Look at the form of the terms. Factorials or $c^n$ suggest the ratio test, an $n$th power suggests the root test, a rational function of $n$ suggests limit comparison with a p-series, and alternating signs suggest the alternating series test. Always check first that the terms go to 0.

### If the terms of a series go to 0, does the series converge?

Not necessarily. The harmonic series $\sum \frac1n$ has terms that approach 0, yet it diverges. Terms going to 0 is required for convergence but is not enough on its own, so the nth-term test can only ever prove divergence.

### What is the difference between absolute and conditional convergence?

A series converges absolutely if $\sum |a_n|$ converges. It converges conditionally if $\sum a_n$ converges but $\sum |a_n|$ diverges. For example, $\sum \frac{(-1)^{n+1}}{n}$ converges conditionally, while $\sum \frac{(-1)^{n+1}}{n^2}$ converges absolutely.

### What happens when the ratio test equals 1?

The test is inconclusive. Both $\sum \frac1n$ (divergent) and $\sum \frac1{n^2}$ (convergent) give a ratio limit of 1, so that value cannot distinguish them. Try a comparison test, the integral test or the p-series rule instead.

### What is the p-series test?

A p-series has the form $\sum_{n=1}^\infty \frac{1}{n^p}$. It converges when $p > 1$ and diverges when $p \le 1$. It is the most common benchmark for comparison tests.