In mathematics, convergence tests are methods of testing for the convergence, conditional convergence, absolute convergence, interval of convergence or divergence of an infinite series $\sum _{n=1}^{\infty }a_{n)$ .

## List of tests

### Limit of the summand

If the limit of the summand is undefined or nonzero, that is $\lim _{n\to \infty }a_{n}\neq 0$ , then the series must diverge. In this sense, the partial sums are Cauchy only if this limit exists and is equal to zero. The test is inconclusive if the limit of the summand is zero. This is also known as the nth-term test, test for divergence, or the divergence test.

### Ratio test

This is also known as d'Alembert's criterion.

Suppose that there exists $r$ such that
$\lim _{n\to \infty }\left|{\frac {a_{n+1)){a_{n))}\right|=r.$ If r < 1, then the series is absolutely convergent. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.

### Root test

This is also known as the nth root test or Cauchy's criterion.

Let
$r=\limsup _{n\to \infty }{\sqrt[{n}]{|a_{n}|)),$ where $\limsup$ denotes the limit superior (possibly $\infty$ ; if the limit exists it is the same value).
If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the root test is inconclusive, and the series may converge or diverge.

The root test is stronger than the ratio test: whenever the ratio test determines the convergence or divergence of an infinite series, the root test does too, but not conversely. For example, for the series

1 + 1 + 0.5 + 0.5 + 0.25 + 0.25 + 0.125 + 0.125 + ... = 4,

convergence follows from the root test but not from the ratio test.[citation needed]

### Integral test

The series can be compared to an integral to establish convergence or divergence. Let $f:[1,\infty )\to \mathbb {R} _{+)$ be a non-negative and monotonically decreasing function such that $f(n)=a_{n)$ . If

$\int _{1}^{\infty }f(x)\,dx=\lim _{t\to \infty }\int _{1}^{t}f(x)\,dx<\infty ,$ then the series converges. But if the integral diverges, then the series does so as well. In other words, the series ${a_{n))$ converges if and only if the integral converges.

#### p-series test

A commonly-used corollary of the integral test is the p-series test. Let $k>0$ . Then $\sum _{n=k}^{\infty }{\bigg (}{\frac {1}{n^{p))}{\bigg ))$ converges if $p>1$ .

The case of $p=1,k=1$ yields the harmonic series, which diverges. The case of $p=2,k=1$ is the Basel problem and the series converges to ${\frac {\pi ^{2)){6))$ . In general, for $p>1,k=1$ , the series is equal to the Riemann zeta function applied to $p$ , that is $\zeta (p)$ .

### Direct comparison test

If the series $\sum _{n=1}^{\infty }b_{n)$ is an absolutely convergent series and $|a_{n}|\leq |b_{n}|$ for sufficiently large n , then the series $\sum _{n=1}^{\infty }a_{n)$ converges absolutely.

### Limit comparison test

If $\{a_{n}\},\{b_{n}\}>0$ , (that is, each element of the two sequences is positive) and the limit $\lim _{n\to \infty }{\frac {a_{n)){b_{n)))$ exists, is finite and non-zero, then $\sum _{n=1}^{\infty }a_{n)$ diverges if and only if $\sum _{n=1}^{\infty }b_{n)$ diverges.

### Cauchy condensation test

Let $\left\{a_{n}\right\)$ be a positive non-increasing sequence. Then the sum $A=\sum _{n=1}^{\infty }a_{n)$ converges if and only if the sum $A^{*}=\sum _{n=0}^{\infty }2^{n}a_{2^{n))$ converges. Moreover, if they converge, then $A\leq A^{*}\leq 2A$ holds.

### Abel's test

Suppose the following statements are true:

1. $\sum a_{n)$ is a convergent series,
2. $\left\{b_{n}\right\)$ is a monotonic sequence, and
3. $\left\{b_{n}\right\)$ is bounded.

Then $\sum a_{n}b_{n)$ is also convergent.

### Absolute convergence test

Every absolutely convergent series converges.

### Alternating series test

Suppose the following statements are true:

• $a_{n)$ are all positive,
• $\lim _{n\to \infty }a_{n}=0$ and
• for every n, $a_{n+1}\leq a_{n)$ .

Then $\sum _{n=k}^{\infty }(-1)^{n}a_{n)$ and $\sum _{n=k}^{\infty }(-1)^{n+1}a_{n)$ are convergent series. This test is also known as the Leibniz criterion.

### Dirichlet's test

If $\{a_{n}\)$ is a sequence of real numbers and $\{b_{n}\)$ a sequence of complex numbers satisfying

• $a_{n}\geq a_{n+1)$ • $\lim _{n\rightarrow \infty }a_{n}=0$ • $\left|\sum _{n=1}^{N}b_{n}\right|\leq M$ for every positive integer N

where M is some constant, then the series

$\sum _{n=1}^{\infty }a_{n}b_{n)$ converges.

### Raabe–Duhamel's test

Let $a_{n}>0$ .

Define

$b_{n}=n\left({\frac {a_{n)){a_{n+1))}-1\right).$ If

$L=\lim _{n\to \infty }b_{n)$ exists there are three possibilities:

• if L > 1 the series converges
• if L < 1 the series diverges
• and if L = 1 the test is inconclusive.

An alternative formulation of this test is as follows. Let { an } be a series of real numbers. Then if b > 1 and K (a natural number) exist such that

$\left|{\frac {a_{n+1)){a_{n))}\right|\leq 1-{\frac {b}{n))$ for all n > K then the series {an} is convergent.

### Bertrand's test

Let { an } be a sequence of positive numbers.

Define

$b_{n}=\ln n\left(n\left({\frac {a_{n)){a_{n+1))}-1\right)-1\right).$ If

$L=\lim _{n\to \infty }b_{n)$ exists, there are three possibilities:

• if L > 1 the series converges
• if L < 1 the series diverges
• and if L = 1 the test is inconclusive.

### Gauss's test

Let { an } be a sequence of positive numbers. If ${\frac {a_{n)){a_{n+1))}=1+{\frac {\alpha }{n))+O(1/n^{\beta })$ for some β > 1, then $\sum a_{n)$ converges if α > 1 and diverges if α ≤ 1.

### Kummer's test

Let { an } be a sequence of positive numbers. Then:

(1) $\sum a_{n)$ converges if and only if there is a sequence $b_{n)$ of positive numbers and a real number c > 0 such that $b_{k}(a_{k}/a_{k+1})-b_{k+1}\geq c$ .

(2) $\sum a_{n)$ diverges if and only if there is a sequence $b_{n)$ of positive numbers such that $b_{k}(a_{k}/a_{k+1})-b_{k+1}\leq 0$ and $\sum 1/b_{n)$ diverges.

## Examples

Consider the series

$\sum _{n=1}^{\infty }{\frac {1}{n^{\alpha ))}.$ (i)

Cauchy condensation test implies that (i) is finitely convergent if

$\sum _{n=1}^{\infty }2^{n}\left({\frac {1}{2^{n))}\right)^{\alpha )$ (ii)

is finitely convergent. Since

$\sum _{n=1}^{\infty }2^{n}\left({\frac {1}{2^{n))}\right)^{\alpha }=\sum _{n=1}^{\infty }2^{n-n\alpha }=\sum _{n=1}^{\infty }2^{(1-\alpha )n)$ (ii) is a geometric series with ratio $2^{(1-\alpha ))$ . (ii) is finitely convergent if its ratio is less than one (namely $\alpha >1$ ). Thus, (i) is finitely convergent if and only if $\alpha >1$ .

## Convergence of products

While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. This can be achieved using following theorem: Let $\left\{a_{n}\right\}_{n=1}^{\infty )$ be a sequence of positive numbers. Then the infinite product $\prod _{n=1}^{\infty }(1+a_{n})$ converges if and only if the series $\sum _{n=1}^{\infty }a_{n)$ converges. Also similarly, if $0 holds, then $\prod _{n=1}^{\infty }(1-a_{n})$ approaches a non-zero limit if and only if the series $\sum _{n=1}^{\infty }a_{n)$ converges .

This can be proved by taking the logarithm of the product and using limit comparison test.