In this tutorial, we review some of the most common tests for the convergence of an infinite series $$ \sum_{k=0}^{\infty} a_k = a_0 + a_1 + a_2 + \cdots $$ The proofs or these tests are interesting, so we urge you to look them up in your calculus text.

Let \begin{eqnarray*} s_0 & = & a_0 \\ s_1 & = & a_1 \\ & \vdots & \\ s_n & = & \sum_{k=0}^{n} a_k \\ & \vdots & \end{eqnarray*} If the sequence $\{ s_n \}$ of partial sums converges to a limit $L$, then the series is said to converge to the sum $L$ and we write

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$$ \sum_{k=0}^{\infty}a_k = L. $$

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For $j \ge 0$, $\sum\limits^{\infty}_{k=0} a_k$ converges if and only if $\sum\limits_{k=j}^{\infty} a_k$ converges, so in discussing convergence we often just write $\sum a_k$.

Example

Consider the geometric series $$ \sum_{k=0}^{\infty} x^k. $$ The $n^{th}$ partial sum is $$ s_n = 1 + x + x^{2} + \cdots + x^{n}. $$ Multiplying both sides by $x$, $$ xs_n = x + x^{2} + x^{3} + \cdots + x^{n+1}. $$ Subtracting the second equation from the first, $$ (1-x)s_n = 1-x^{n+1}, $$ so for $x \not= 1$, $$ s_n = \frac{1-x^{n+1}}{1-x}. $$ For $|x| < 1$, $$ \lim_{n \rightarrow \infty} s_n = \frac{1}{1-x}. $$ It is easy to see that $\sum\limits_{k=0}^{\infty}x^{k}$ diverges for $|x| \ge 1$. Thus $\sum\limits_{k=0}^{\infty}x^{k} = \frac{1}{1-x}$ for $|x| < 1$ and diverges for $|x| \ge 1$.

Divergence Test

If $\lim\limits_{k \rightarrow \infty} a_k \not= 0$, then $\sum\limits_{k=0}^{\infty}a_k$ diverges.

Example

The series $\sum\limits_{k=0}^{\infty}\frac{k}{2k+1}$ diverges, since $\lim\limits_{k \rightarrow \infty} \frac{k}{2k+1} = 1/2 \not= 0$.

Integral Test

Let $f(x)$ be continuous, decreasing, and positive for $x \ge 1$. Then $\sum\limits_{k=1}^{\infty}f(k)$ converges if and only if $\int\limits_{1}^{\infty} f(x)dx$ converges.

Example

Consider the p-series $$ \sum_{k=1}^{\infty}\frac{1}{k^p} = \frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + \cdots $$ Since $$ \int_{1}^{\infty}\frac{1}{x^{p}}dx = \left\{\begin{array}{l@{,\quad}l} \left.\frac{1}{1-p}x^{1-p}\right|_{1}^{\infty} & p>1 \\ \left.\ln |x| \right|_{1}^{\infty} & p=1 \\ \left.\frac{1}{1-p}x^{1-p}\right|_{1}^{\infty} & 0 < p < 1 \end{array} \right. = \left\{ \begin{array}{c} \frac{1}{1-p}\\ \infty \\ \infty, \end{array}\right. $$ the series converges for $p>1$ and diverges for $0 < p \le 1$.

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The divergent p-series $$ \sum_{k=1}^{\infty}\frac{1}{k} $$ with $p=1$ is called the Harmonic Series.

Comparison Test

Let $\sum a_k$ and $\sum b_k$ be series with non-negative terms. If $a_k \le b_k$ for all $k$ sufficiently large, then

1. If $\sum b_k$ converges, then $\sum a_k$ also converges.
2. If $\sum a_k$ diverges, then $\sum b_k$ also diverges.

Examples

Since $\sum\limits_{k=1}^{\infty}\frac{1}{k^2}$ converges, so does $\sum\limits_{k=1}^{\infty}\frac{1}{k^2 + 3}$. Since $\sum\limits_{k=1}^{\infty}\frac{1}{k}$ diverges, so does $\sum\limits_{k=1}^{\infty}\frac{1}{\ln |k+1|}$.

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$\frac{1}{k^2 +3} < \frac{1}{k^2}$ for all $k$. $\frac{1}{\ln |k+1|} > \frac{1}{k}$ for $k \ge 2$.

Limit Comparison Test

Let $\sum a_k$ and $\sum b_k$ be series with positive terms. If $$ \lim_{k \rightarrow \infty}\frac{a_k}{b_k} = L $$ where $0 < L < \infty$ then $\sum a_k$ and $\sum b_k$ either both converge or both diverge.

Example

The series $\sum\limits_{k=1}^{\infty}\frac{k^2-1}{5k^3}$ diverges, since $\sum\limits_{k=1}^{\infty}\frac{1}{k}$ diverges and $$ \lim_{k \rightarrow \infty} \frac{\frac{k^2-1}{5k^3}}{\frac{1}{k}} = \lim_{k \rightarrow \infty} \frac{k^2-1}{5k^2} = \frac{1}{5}. $$

Ratio Test

Let $\sum a_k$ be a series with positive terms and suppose that $$ \lim_{k \rightarrow \infty}\frac{a_{k+1}}{a_k} = L. $$

1. If $L < 1$, then $\sum a_k$ converges.
2. If $L > 1$, then $\sum a_k$ diverges.
3. If $L = 1$, then the test is inconclusive.

Example

The series $\sum\limits_{k=1}^{\infty}\frac{1}{k!}$ converges, since $$ \lim_{k \rightarrow \infty}\frac{\frac{1}{(k+1)!}}{\frac{1}{k!}} = \lim_{k \rightarrow \infty}\frac{1}{k+1} = 0. $$

Root Test

Let $\sum a_k$ be a series with non-negative terms and suppose that $$ \lim_{k \rightarrow \infty} (a_k)^{\frac{1}{k}} = L. $$

1. If $L < 1$, then $\sum a_k$ converges.
2. If $L > 1$, then $\sum a_k$ diverges.
3. If $L = 1$, then the test is inconclusive.

Example

The series $\sum\limits_{k=0}^{\infty} \left( \frac{k}{2k+1} \right)^{k}$ converges, since $$ \lim_{k \rightarrow \infty} \left[\left(\frac{k}{2k+1}\right)^{k}\right]^{\frac{1}{k}} = \lim_{k \rightarrow \infty} \frac{k}{2k+1} = \frac{1}{2}. $$

Alternating Series Test

Consider the alternating series $$ \sum_{k=0}^{\infty}(-1)^{k}a_k $$ where $a_k > 0$ for all $k \ge 0$.

If $a_{k+1} < a_k$ for all $k$ and $\lim\limits a_k = 0$, then $\sum\limits_{k=0}^{\infty}(-1)^{k}a_k$ converges.

Example

The series $\sum\limits_{k=0}^{\infty} \frac{(-1)^{k}}{k+1}$ converges, since $\frac{1}{(k+1) + 1} < \frac{1}{k+1}$ and $\lim\limits_{k \rightarrow \infty}\frac{1}{k+1} =0$. This series is conditionally convergent, rather than absolutely convergent, since $\sum\limits_{k=0}^{\infty}\left|\frac{(-1)^{k}}{k+1}\right| = \sum\limits_{k=0}^{\infty}\frac{1}{k+1}$ diverges.

For a particular series, one or more of the common convergence tests may be most convenient to apply.