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Leibniz Test

Definition

If the alternating series $\sum_{n=1}^{\infty} (-1)^{n-1} a_n$ satisfies:

then the series converges.

Note: This is a sufficient condition. Alternating series that satisfy these conditions must converge.

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$\sum$	Greek letter	Sigma	Summation symbol, representing series
$\infty$	Mathematical symbol	Infinity	Represents infinite series, infinite number of terms
$\lim$	Mathematical symbol	Limit	Represents limit of sequence or function

Sufficient conditions for the convergence of alternating series $\sum_{n=1}^{\infty} (-1)^{n-1} a_n$ :

Determine the convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ .

Solution: This is an alternating series, where $a_n = \frac{1}{n}$

The series satisfies the conditions of the Leibniz test, so it converges.

Determine the convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{1/2}}$ .

Reference Answer

series convergenceconditional convergencealternating seriesLeibniz test

Approach: First check for absolute convergence. If the series converges absolutely, then the original series converges.

Steps:

Consider the absolute value series: $\sum_{n=1}^{\infty} \frac{1}{n^{1/2}}$
This is a p-series with $p = \frac{1}{2} \leq 1$ , so it diverges.
Since the absolute value series diverges, we need to check further.
The original series is an alternating series with $a_n = \frac{1}{n^{1/2}}$ .
Check the Leibniz test conditions:
- $a_n > 0$ ✓
- $a_{n+1} < a_n$ ✓ (because $\frac{1}{(n+1)^{1/2}} < \frac{1}{n^{1/2}}$ )
- $\lim_{n \to \infty} a_n = 0$ ✓
The series satisfies the Leibniz test conditions, so it converges.

Answer: The series converges (conditional convergence).

Symbol	Type	Pronunciation/Explanation	Meaning in This Article
$a_n$	Mathematical symbol	General term	The nth term in the series

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