# nth-Term Test for Divergence: Statement, Examples and Limits

> The nth-term test: if the terms of a series do not approach 0, the series diverges. The statement, why it cannot prove convergence, and examples.

Source: https://practicecalculus.com/rules/nth-term-test
Updated: 2026-11-03

$$ \lim_{n\to\infty} a_n \ne 0 \;\Longrightarrow\; \sum a_n \text{ diverges} $$

## At a glance

- **Also called:** The divergence test, the term test
- **Can prove:** Divergence only
- **Cannot prove:** Convergence. If $a_n \to 0$, the test is inconclusive
- **Classic example:** $\sum \frac{1}{n}$ has $a_n \to 0$ but still diverges
- **Use it:** First, before any other convergence test

## Key takeaways

- If $\lim a_n \ne 0$ or the limit does not exist, $\sum a_n$ diverges.
- If $\lim a_n = 0$, the nth-term test says nothing; use another test.
- Always check this first: it is the fastest way to rule a series out.

The **nth-term test** (or divergence test) is the quickest check in the study of series. A series can only add up to a finite number if its terms shrink to zero, so if they do not, the series diverges.

## The statement

For any series $\sum a_n$:

$$\lim_{n\to\infty} a_n \ne 0 \quad\text{or the limit does not exist} \quad\Longrightarrow\quad \sum_{n=1}^{\infty} a_n \text{ diverges}$$

The equivalent form is a necessary condition for convergence:

$$\sum a_n \text{ converges} \quad\Longrightarrow\quad \lim_{n\to\infty} a_n = 0$$

The converse is **false**. Terms going to zero is necessary, not sufficient.

## Why it works

Let $S_n$ be the $n$th partial sum. Then $a_n = S_n - S_{n-1}$. If the series converges to $S$, both $S_n$ and $S_{n-1}$ approach $S$, so

$$a_n = S_n - S_{n-1} \to S - S = 0$$

## Reading the result

| Value of $\lim a_n$ | Conclusion |
| --- | --- |
| A nonzero number | Diverges |
| $\pm\infty$ | Diverges |
| Does not exist | Diverges |
| $0$ | **Inconclusive**: use another test |

```graph
title: 'Terms aₙ = n/(2n+1) approach 1/2, while 1/n approaches 0'
caption: 'The terms $\frac{n}{2n+1}$ level off at $\frac{1}{2}$, so their series diverges. The terms $\frac{1}{n}$ do go to $0$, but that alone does not make $\sum \frac{1}{n}$ converge.'
x: [0, 12]
y: [0, 1.1]
functions:
  - expr: x/(2x+1)
    label: y = x/(2x+1)
    domain: [1, 12]
  - expr: 1/x
    label: y = 1/x
    domain: [1, 12]
    dashed: true
  - expr: '0.5'
    label: y = 1/2
    dashed: true
```

## Worked examples

**Example 1: a rational expression.** For $\sum \frac{n}{2n+1}$:

$$\lim_{n\to\infty} \frac{n}{2n+1} = \frac{1}{2} \ne 0$$

The series diverges. You are adding numbers close to $\frac{1}{2}$ forever.

**Example 2: oscillating terms.** For $\sum (-1)^n$, the terms alternate between $1$ and $-1$, so $\lim a_n$ does not exist. The series diverges.

**Example 3: a trig term.** For $\sum \cos\left(\frac{1}{n}\right)$, $\frac{1}{n} \to 0$ and $\cos 0 = 1$, so $a_n \to 1$. The series diverges.

**Example 4: a famous limit.** For $\sum \left(1 + \frac{1}{n}\right)^n$, the terms approach $e \ne 0$. The series diverges.

**Example 5: inconclusive.** For $\sum \frac{1}{n^2}$ and $\sum \frac{1}{n}$, the terms go to $0$ in both cases. The test says nothing. In fact $\sum \frac{1}{n^2}$ converges and $\sum \frac{1}{n}$ diverges, which other tests (the integral test or the $p$-series rule) reveal.

## Divergence vs convergence at a glance

Think of the test as a gate. Every convergent series passes through it, because its terms must shrink to zero. Many divergent series fail at the gate, but some, like the harmonic series, slip through and only reveal their divergence under a stronger test. So a failed check is a final answer, while a passed check only means the real work is still ahead. This one-way logic is the single most important thing to remember about the test.

## Common mistakes

1. **Concluding convergence from $a_n \to 0$.** This is the most common series mistake. The harmonic series is the counterexample.
2. **Taking the limit of the partial sums instead of the terms.** The test uses $a_n$, the individual terms.
3. **Forgetting the "does not exist" case.** Oscillating terms like $(-1)^n$ or $\sin n$ also fail to approach $0$.
4. **Skipping it.** Running the [ratio test](/rules/ratio-test) on $\sum \frac{n}{2n+1}$ gives a limit of $1$ and no answer, while the nth-term test settles it at once.

## Where it fits

Use the nth-term test as the first step for every series. If the terms do go to zero, move on: try the [geometric series formula](/rules/geometric-series-formula) for constant ratios, the [alternating series test](/rules/alternating-series-test) for alternating signs, and the [ratio test](/rules/ratio-test) for factorials and exponentials. The full guide to [convergence tests](/blog/convergence-tests) shows how to choose between them.

## Frequently asked questions

### What does the nth-term test say?

If the terms $a_n$ of a series do not approach $0$, the series $\sum a_n$ diverges. It is a test for divergence only.

### Can the nth-term test show that a series converges?

No. If $a_n \to 0$ the test is inconclusive. The harmonic series $\sum \frac{1}{n}$ has terms going to $0$ but diverges.

### What if the limit of a_n does not exist?

Then the series diverges as well. For example $\sum (-1)^n$ has terms that alternate between $1$ and $-1$, so its terms do not approach $0$.

### Why is the nth-term test true?

If $\sum a_n$ converges to $S$, then $a_n = S_n - S_{n-1} \to S - S = 0$. So convergence forces the terms to vanish, and terms that do not vanish rule out convergence.