Strategy for Series

Main idea

No single test handles every series, but each test has a kind of term it's built for. Look at what the terms are made of first, and that usually tells you which test to use within a few seconds.

Explanation

A series strategy is simply a habit for picking the right test, so you don't waste time trying all of them. You now have a whole toolbox: the divergence test, geometric and p-series rules, the integral test, comparison, alternating series, ratio and root. Each one is good at a particular shape of term, the way a wrench fits bolts and a screwdriver fits screws. The table below looks like a lot, but every row is just two pieces: what the terms look like (the clue) and which test that shape points to (the tool). Once you've seen each shape a couple of times, it gets quick; bear with it.

Work down this list and stop at the first test that fits.

If the terms look like	Try
they don't approach 0	Divergence test: it diverges
$\dfrac{1}{n^p}$ or $ar^n$	p-series or geometric rules
a rational or algebraic function of $n$	Limit comparison with $\frac{1}{n^p}$ , keeping the highest powers
$(-1)^n b_n$	Check $\sum b_n$ for absolute convergence, then the alternating series test
factorials, or powers like $5^n$ and $n^n$	Ratio test
something raised to the $n$ th power	Root test
$f(n)$ with an easy antiderivative, like $\dfrac{1}{n\ln n}$	Integral test

A few habits make this faster:

Always try the divergence test first. It takes five seconds and ends many problems.
Think about sizes. $\ln n \ll n^p \ll a^n \ll n! \ll n^n$ for large $n$ . The fastest-growing piece in the numerator and denominator decides what the series looks like, since everything else becomes a rounding error far out.
Signs. If the terms have mixed signs, decide absolute convergence first. It's often easier, and if it works you get the stronger result for free.

A common mistake. It's tempting to settle convergence by adding up a bunch of terms on a calculator and seeing whether the total looks like it's leveling off. That can fool you badly. There's an animation further down showing $\sum \frac1n$ , which diverges, and $\sum \frac{1}{n^{1.1}}$ , which converges, looking almost identical for the first 200 terms. Only a test can tell them apart.

See it

Running totals of Σ 1/n, Σ 1/n^1.1 and Σ 1/n² over the first 200 terms. Σ 1/n² levels off almost immediately near 1.64. The other two look nearly alike: after 200 terms they're both still creeping upward, at about 5.9 and 4.7. Yet Σ 1/n never stops, while Σ 1/n^1.1 eventually stops at about 10.58, far off the right edge. You can't see the difference here, which is exactly why we need the tests.

Worked examples

Example 1

Classify $\displaystyle\sum \frac{n^2 - 1}{n^3 + 4n}$ .

1
Start with the divergence test: the terms go to 0, so no quick verdict there.

Example 2

Classify $\displaystyle\sum \frac{(-1)^n\, 3^n}{n!}$ .

1
The factorial is the clue: use the ratio test on the absolute values. The ratio is $\frac{3}{n+1} \to 0$ .

Absolutely convergent. (Its sum is $e^{-3}$ .)

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Classify $\displaystyle\sum \frac{n^2 + 1}{2n^2 - 3}$ .

Problem 2+5 XP

Classify $\displaystyle\sum n^3 e^{-n}$ .

Problem 3+5 XP

Classify $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{\ln n}$ .

Problem 4+5 XP

Classify $\displaystyle\sum \frac{1}{n\sqrt n}$ .

Problem 5+5 XP

Classify $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n\ln n}$ .

Problem 6+5 XP

Find $\displaystyle\sum_{n=0}^{\infty} \frac{2^n + 1}{3^n}$ .

Problem 7+5 XP

You flip a fair coin until it lands heads. The chance it takes exactly $n$ flips is $\frac{1}{2^n}$ , so the average number of flips is $\sum_{n=1}^\infty\frac{n}{2^n}$ . Decide which test shows it converges, then find its value.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Classify $\displaystyle\sum_{n=1}^\infty\frac{1}{n^{1 + 1/n}}$ . (Careful: every exponent is bigger than $1$ .)

Where it's used

Choosing a test is a small version of everyday engineering: recognize the structure of a problem before reaching for a tool.

Retrying Packets on a Lossy LinkCalc II · Computer Science Why a Square Wave Has Finite PowerCalc II · Electrical Engineering

The Root Test Estimating the Value of a Series