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

# 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.
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 $\boxed{\text{divergent}}$

# 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. $\text{The ratio test is inconclusive if } \lim_{n \to \infty}{\sqrt[n]{a_n}} = 1$

# Examples

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