Series Convergence Tests, with a Strategy for Choosing

Series, partial sums and convergence

An infinite series is a sum of infinitely many terms, $a_1 + a_2 + a_3 + \cdots$, written $\sum a_n$. You cannot add infinitely many numbers directly, so instead you add the first $N$ of them to get the partial sum $S_N$, and ask what happens as $N$ grows. If the partial sums approach a finite limit $S$, the series converges and its sum is $S$. Otherwise it diverges.

Adding infinitely many positive numbers can give a finite total. The series $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$ has partial sums $0.5$, $0.75$, $0.875$ and so on, each one halfway closer to 1, so the sum is exactly 1. Picture walking towards a wall, each step covering half the remaining distance: you take infinitely many steps but never travel more than the original distance. The whole subject rests on limits, applied to the sequence of partial sums.

Geometric series

A geometric series multiplies by the same ratio $r$ each time. It is one of the few kinds of series whose sum you can write down exactly:

If $|r| \ge 1$, the terms do not shrink to zero and the series diverges. For example, $3 + 1 + \frac{1}{3} + \frac{1}{9} + \cdots$ has $a = 3$ and $r = \frac{1}{3}$, so its sum is $\frac{3}{2/3} = 4.5$. Repeating decimals are geometric series in disguise: $0.999\ldots = \frac{9}{10} + \frac{9}{100} + \cdots = \frac{0.9}{1 - 0.1} = 1$.

The divergence test

If the terms $a_n$ do not approach 0, the series diverges. This should always be your first check, because it takes seconds. For example, $\sum\frac{n}{n + 1}$ diverges since its terms approach 1, so you keep adding almost 1 forever.

The converse is false, and this is the most common misconception about series. Terms approaching zero do not guarantee convergence. The harmonic series $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots$ has terms that shrink to 0, yet it diverges. Group its terms: $\frac{1}{3} + \frac{1}{4} > \frac{1}{2}$, the next four terms exceed $\frac{1}{2}$, the next eight exceed $\frac{1}{2}$, and so on. Adding infinitely many halves gives an unbounded total. The growth is very slow, though: the first million terms add up to only about 14.4.

p-series and the integral test

A p-series has the form $\sum\frac{1}{n^p}$. Its behaviour depends only on $p$:

So $\sum\frac{1}{n^2}$ converges, and its sum is famously $\frac{\pi^2}{6} \approx 1.6449$, while $\sum\frac{1}{\sqrt{n}}$ diverges. The proof uses the integral test: if $f$ is positive, continuous and decreasing with $f(n) = a_n$, then $\sum a_n$ and $\int_1^{\infty} f(x)\,dx$ either both converge or both diverge. Since $\int_1^{\infty}\frac{dx}{x^2} = 1$ is finite and $\int_1^{\infty}\frac{dx}{x}$ is infinite, the two key cases follow. Those improper integrals are worked out in definite vs indefinite integrals.

Comparison tests

Comparison tests decide a series by relating it to one you already know, usually a geometric or p-series. For series with positive terms:

• Direct comparison: if $0 \le a_n \le b_n$ and $\sum b_n$ converges, so does $\sum a_n$. If $a_n \ge b_n \ge 0$ and $\sum b_n$ diverges, so does $\sum a_n$.
• Limit comparison: if $\lim\frac{a_n}{b_n}$ is a positive finite number, the two series behave the same way.

To pick a comparison series, keep only the highest power of $n$ in the numerator and in the denominator. What remains tells you which p-series or geometric series to compare with.

The ratio and root tests

The ratio test looks at how each term compares with the one before. Let $L = \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|$. If $L < 1$ the series converges absolutely, if $L > 1$ it diverges, and if $L = 1$ the test gives no information. It is the natural choice for series containing factorials or powers such as $2^n$.

The root test uses $L = \lim\sqrt[n]{|a_n|}$ with the same conclusions, and suits terms raised to the power $n$. For $\sum\left(\frac{n}{2n + 1}\right)^n$, the $n$th root is $\frac{n}{2n + 1} \to \frac{1}{2}$, so the series converges.

Alternating series

An alternating series switches sign each term. The alternating series test says that if the absolute values of the terms decrease and approach 0, the series converges. The alternating harmonic series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ therefore converges, and its sum is $\ln 2 \approx 0.6931$.

There is a bonus: the error after stopping at any term is no larger than the first term left out. So stopping the alternating harmonic series after 99 terms gives $\ln 2$ to within $\frac{1}{100}$.

A series converges absolutely if $\sum|a_n|$ converges, and conditionally if it converges only thanks to the cancelling signs. The alternating harmonic series is conditionally convergent, since without the signs it is the divergent harmonic series. Absolute convergence is the stronger property: rearranging the terms of an absolutely convergent series never changes its sum, while a conditionally convergent series can be rearranged to add up to any number you like.

A strategy for choosing a test

1. Do the terms approach 0? If not, the series diverges.
2. Is it geometric or a p-series? Apply the known rule.
3. Does it contain factorials or $n$th powers like $3^n$? Use the ratio test.
4. Is the whole term raised to the power $n$? Use the root test.
5. Is it a rational function of $n$ or similar? Use a limit comparison with a p-series.
6. Do the signs alternate? Use the alternating series test, then check absolute convergence.
7. Is the term a function you can integrate easily? Consider the integral test.

The series convergence calculator applies these tests, reports which one decides the question and estimates the sum. Power series, whose terms contain $x$, are handled the same way: the ratio test gives a radius of convergence, and the endpoints are checked separately, which is how the intervals in our Taylor series guide are found.

Telescoping series and practice problems

A telescoping series is one whose partial sums collapse because neighbouring terms cancel. Writing $\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$ shows that the partial sum $S_N$ equals $1 - \frac{1}{N + 1}$, which approaches 1. Spotting a telescoping pattern, often with partial fractions, gives the exact sum instead of only a yes or no answer.

Test yourself on these series before checking the answers: $\sum\frac{5}{4^n}$ is geometric with ratio one quarter and converges; $\sum\frac{n^2}{n^2 + 1}$ diverges by the divergence test because its terms approach 1; $\sum\frac{1}{n^{1.01}}$ converges as a p-series with $p > 1$; $\sum\frac{3^n}{n!}$ converges by the ratio test; and $\sum\frac{(-1)^n}{\sqrt{n}}$ converges conditionally, since the alternating series test applies but the series of absolute values is a divergent p-series.

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