is created by David Witten, a mathematics and computer science student at Vanderbilt University. For more information, see the "About" page.

Testing for Convergence and Divergence

Determining Whether a Series is Convergent or Divergent

If {$a_n$} is an infinite sequence, then $$\sum_{n = 1}^{\infty} a_n = a_1 + a_2 + ... a_n + ...$$ One way to find the sum of an infinite series is to consider the sequence of partial sums. $$S_1 = a_1$$ $$S_2 = a_1 + a_2$$ $$S_n = a_1 + a_2 + a_3 + ... + a_n$$ If the sequence of partial sums ${S_n}$ converges to S, then the seres converges. S is the sum of the series.

That doesn't really help though, we want to know whether a series converges or diverges, and finding the limit of partial sums may get difficult. Here are some methods to actually solve convergent series:

Telescoping Series

Let's say you have this: $$\sum_{n = 1}^{\infty} \frac{1}{n} - \frac{1}{n + 1}$$ $$\text{That equals: } (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + ...$$ $$=1 + (-\frac{1}{2} + \frac{1}{2}) + (-\frac{1}{3} + \frac{1}{3}) + (-\frac{1}{4} + \frac{1}{4}) - \frac{1}{5} ... $$ Notice that you can cross out the $\frac{1}{2}$s, the $\frac{1}{3}$s, and everything except for the 1. So the partial sum becomes $1 - \frac{1}{n+1}$. $$\lim_{n \to \infty} {1 - \frac{1}{n + 1}} = 1$$ So, that telescoping series equals 1.

Geometric Series

Generally a geometric series looks like this: $$\sum_{n = 1}^{\infty} ar^n = a + ar + ar^2 + ... + ar^n$$

A geometric series with ratio r diverges if |r| >= 1. If 0 < |r| < 1, then the series converges to the sum

$$\sum_{n = 0}^{\infty} ar^n = \frac{a}{1-r}, 0 < |r| < 1$$ $$\text{For example, let's say you have this:}$$ $$\sum_{n = 0}^{\infty} 3^{n+2}9^{-n+2}$$ $$\text{It becomes: } (3^2)(9^2)\sum_{n = 0}^{\infty}\frac{3^n}{9^n}$$ $$\text{Which equals: } 729 \sum_{n = 0}^{\infty} (\frac{1}{3})^n$$ $$= \frac{729}{1 - \frac{1}{3}} = 1093.5$$

N-th Term Test

This has no prerequisites, and it should be the first test that anyone does. $$\lim_{n \to \infty} a_n \ne 0 \implies \sum_{n=1}^{\infty}a_n \text{ diverges}.$$ Be careful, because if it does equal 0, that doesn't imply convergence. Take $\sum{\frac{1}{n}}$, the $\lim_{n \to \infty}{\frac{1}{n}} = 0$, but it still diverges (we'll see why later).

Integral Test


This test has a few prerequisites. If f is positive, continuous, and decrease for $x \geq M$ (doesn't have to be 1), then $\sum_{n = 1}^{\infty}a_n$ and $\int_{1}^{\infty}\mathrm{f(x)}\, \mathrm{d}x$ both converge or diverge. Another way of writing this is the sum converges f and only if the integral converges.
You can think about this like the sum from n = 1 was a LHS, and the sum from n = 2 was a RHS, and if the integral converges, the sum must also converge. If the integral diverges, then the sum must also diverge. Example: $$\sum_{n = 0}^{\infty}{\frac{1}{n\ln(n)}}$$ $$\int_0^{\infty}\mathrm{\frac{1}{x\ln(x)}}\, \mathrm{d}x$$ $$\int\mathrm{\frac{1}{u}}\, \mathrm{d}u \text{ ,u = ln(x)}$$ $$=\frac{1}{\ln(x)}|_{0}^{\infty}$$ This doesn't diverge, so neither does the series. Therefore, it's $\boxed{\text{divergent}}$

Example: $$\int_{2}^{\infty} \dfrac{1}{\sqrt{4x^2 + 1}}\,\mathrm{d}x$$ Because the integral test is if and only if, we can see if the sum is convergent. $$\sum_{n = 2}^{\infty} \dfrac{1}{\sqrt{4n^2 + 1}}$$ We now do a limit comparison test (described below) $$\lim_{n \to \infty} \dfrac{n}{\sqrt{4n^2 + 1}} = \dfrac{2n}{2\sqrt{4n^2 + 1}} = \dfrac{\sqrt{4n^2}}{2\sqrt{4n^2 + 1}} = \dfrac{\sqrt{4n^2 + 1 - 1}}{2\sqrt{4n^2 + 1}} = \dfrac{1}{2}\sqrt{1 - \dfrac{1}{4n^2 + 1}}$$ $$lim_{n \to \infty} \dfrac{1}{2}\sqrt{1 - \dfrac{1}{4n^2 + 1}}= \dfrac{1}{2}$$ Now, we know that this converges iff $\sum\dfrac{1}{n}$ converges. This diverges (by the p-series test), so we know that our original integral diverges as well.

P-Series Test

This is an extension of the Integral Test. This is a p-series: $$\sum_{n=1}^{\infty}\frac{1}{n^p} = \frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + ...$$ 1.If $p > 1$, the p-series converges.
2. If $0 < p \leq 1$, the p-series diverges.

Direct/Basic Comparison Test

As the name says, it is a basic comparison test. Basically, if the function is between 0 and a convergent function, it's convergent. If it's greater than a divergent function, it diverges. $$\text{Let} 0 < a_n \leq b_n \text{for all n}$$ $$\text{If} \sum_{n=1}^{\infty}b_n \text{ converges, then } \sum_{n=1}^{\infty}a_n \text{ converges} $$ $$\text{If} \sum_{n=1}^{\infty}a_n \text{ diverges, then } \sum_{n=1}^{\infty}b_n \text{ diverges} $$

Limit Comparison Test

Suppose $a_n$ > 0, $b_n$ > 0, and $$lim_{n \to \infty} {\frac{a_n}{b_n}} = L$$ $$\text{where L is finite and positive (non-zero and non-infinite). Then the two series } \sum{a_n} \text{and } \sum{b_n} \text{ both converge or both diverge.}$$

Alternating Series

$$\text{An alternating series is one of the form: } \sum_{n=1}^{\infty}(-1)^na_n$$

There are two things you can do for alternating series

Test for Absolute Convergence 

If the series $\sum{|a_n|}$ converges, then the series $\sum{a_n}$ also converges. This is useful for $\sin(n) \text{ or } \cos(n) * a_n$ If that works, it's absolutely convergent.

If that fails, you go on to

Alternating Series Test

The alternating series $\sum_{n=1}^{\infty}{(-1)^na_n}$ converge IF $\lim_{n \to \infty} a_n = 0$ and $a_{n+1} < a_n \forall n$
If that works, but the test for absolute convergence fails, then it's conditionally convergent, meaning it's convergent when there are negative terms, and it's divergent when it's always positive.

Alternating Series Remainder

This isn't a test, it's more of an extra.

If a convergent alternating series satisfies the condition $a_{n+1}< a_n$, then the absolute value of $R_n$ (the remainder) is less than (or equal to) the next term. $$|S - S_n| = |R_n| \leq a_{n+1}$$


Ratio Test

Let $\sum{a_n}$ be a series with nonzero terms.
1. $\sum{a_n}$ converges absolutely if $\lim_{n \to \infty}{|\frac{a_{n+1}}{a_n}|} < 1$
2. $\sum{a_n}$ diverges if $\lim_{n \to \infty}{|\frac{a_{n+1}}{a_n}|} > 1 \text{ or } = \infty$
3. $\text{The ratio test is inconclusive if } \lim_{n \to \infty}{|\frac{a_{n+1}}{a_n}|} = 1$

Root Test

Let $\sum{a_n}$ be a series.
1. $\sum{a_n}$ converges absolutely if $\lim_{n \to \infty}{\sqrt[n]{a_n}} < 1$
2. $\sum{a_n}$ diverges if $\lim_{n \to \infty}{\sqrt[n]{a_n}} > 1 \text{ or } = \infty$
3. $\text{The ratio test is inconclusive if } \lim_{n \to \infty}{\sqrt[n]{a_n}} = 1$


$$\sum_{n=1}^{\infty}(-1)^n\frac{(\frac{3}{2})^n}{n^2} \text{: Converge or Diverge?}$$ $$\text{Test for absolute convergence: }$$ $$\sum_{n=1}^{\infty}\frac{(\frac{3}{2})^n}{n^2}$$ $$\text{Ratio test:} \lim_{n \to \infty}{\frac{\frac{(3/2)^{n+1}}{(n+1)^2}}{\frac{(3/2)^n}{n^2}}}$$ $$=\lim_{n \to \infty}{\frac{3}{2} * \frac{n^2}{n^2 + 2n + 1}}=\frac{3}{2} \therefore{} \text{Not absolutely convergent}$$ $$\text{Alternating Series Test: } \lim_{n \to \infty}{\frac{(3/2)^n}{n^2}}$$ $$=\text{(L'H) }\lim_{n \to \infty}{\frac{(3/2)^n \ln{3/2}}{2n}}=\text{(L'H) }\lim_{n \to \infty}{\frac{(3/2)^n (\ln{3/2})^2}{2}} > 1 \therefore{} \text{ Diverge} $$
$$\sum_{n=2}^{\infty}\frac{1}{nln(n)}$$ $$\text{Integral Test: } \int_{2}^{\infty}{\frac{1}{xln{x}}}\, \mathrm{d}x$$ $$u = \ln{x}, \int_{ln(2)}^{\infty}{\frac{1}{u}}\,\mathrm{d}u$$ $$ln(x)|_{ln(2)}^{\infty} = \infty \therefore{} Diverge$$
$$\sum_{n=1}^{\infty}\frac{\ln{n}}{n^{3/2}}$$ $$\text{You can prove that }\ln{n} < n^a \text{ } \forall a > 0$$ $$\text{Proof: } \lim_{n \to \infty}{\frac{n^a}{\ln{n}}} = \text{(L'H)} \lim_{n \to \infty}{an^{a-1} * n} = lim_{n \to \infty}{an^a} = \infty$$ $$\therefore{} n^a > \ln{a} \text{ after some point M}$$ $$\text{So, } \frac{\ln{n}}{n^{3/2}} < \frac{n^{1/10}}{n^{3/2}}$$ $$ \frac{n^{1/10}}{n^{3/2}} = \frac{1}{n^{1.4}} \text{ which converges, by the p-series test, } p = 1.4$$ $$\text{By the Basic Comparison Test, } \sum_{n=1}^{\infty}\frac{\ln{n}}{n^{3/2}} \boxed{\text{ converges}}$$
David Witten

Integrals with Functions as Bounds

Separable Differential Equations