Convergence tests

If the limit of the summand is undefined or nonzero, that is ${\displaystyle \lim _{n\to \infty }a_{n}\neq 0}$, 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 d'Alembert's criterion.

Suppose that there exists ${\displaystyle r}$ such that

${\displaystyle \lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|=r.}$

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.

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

Let

${\displaystyle r=\limsup _{n\to \infty }{\sqrt[{n}]{|a_{n}|}},}$

where ${\displaystyle \limsup }$ denotes the limit superior (possibly ${\displaystyle \infty }$; 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.[1] 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.

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

$${\displaystyle \int _{1}^{\infty }f(x)\,dx=\lim _{t\to \infty }\int _{1}^{t}f(x)\,dx<\infty ,}$$

then the series converges. But if the integral diverges, then the series does so as well. In other words, the series ${\displaystyle {a_{n}}}$ converges if and only if the integral converges.

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

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

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

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

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

Suppose the following statements are true:

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

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

Every absolutely convergent series converges.

Suppose the following statements are true:

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

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

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

• ${\displaystyle a_{n}\geq a_{n+1}}$

• ${\displaystyle \lim _{n\rightarrow \infty }a_{n}=0}$

• ${\displaystyle \left|\sum _{n=1}^{N}b_{n}\right|\leq M}$ for every positive integer N

where M is some constant, then the series

${\displaystyle \sum _{n=1}^{\infty }a_{n}b_{n}}$

converges.

Let ${\displaystyle a_{n}>0}$.

Define

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

If

${\displaystyle L=\lim _{n\to \infty }b_{n}}$

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 { a_n } be a series of real numbers. Then if b > 1 and K (a natural number) exist such that

${\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|\leq 1-{\frac {b}{n}}}$

for all n > K then the series {a_n} is convergent.

Let { a_n } be a sequence of positive numbers.

Define

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

If

${\displaystyle L=\lim _{n\to \infty }b_{n}}$

exists, there are three possibilities:[2][3]

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

Let { a_n } be a sequence of positive numbers. If ${\displaystyle {\frac {a_{n}}{a_{n+1}}}=1+{\frac {\alpha }{n}}+O(1/n^{\beta })}$ for some β > 1, then ${\displaystyle \sum a_{n}}$ converges if α > 1 and diverges if α ≤ 1.[4]

• For some specific types of series there are more specialized convergence tests, for instance for Fourier series there is the Dini test.