Series: comparison, ratio, root, Raabe, condensation, alternating, rearrangements

In one sentence

Choosing the right test takes ten seconds if you recognise the shape of the terms, and the questions are built almost entirely around the handful of implications that look true and are not.

Why the exam asks it

Because a series question can pack four independent judgements into one stem, and because the gap between converges and converges absolutely is a clean, un-guessable line. Conditional convergence is where the marks are lost, and the examiner knows it.

The idea

Choosing a test

Start with the free check: if $a_n \not\to 0$, the series diverges. No test needed. The converse is famously false — $\sum 1/n$ has terms going to zero and diverges anyway — and that false converse is the most reused distractor in the topic.

For positive terms, let the shape choose:

• Looks like $1/n^p$ or a geometric series → comparison, or limit comparison. Remember $\sum 1/n^p$ converges exactly when $p > 1$.
• Factorials or $n^n$ → ratio test. If $a_{n+1}/a_n \to L$: converges for $L < 1$, diverges for $L > 1$, and tells you nothing at $L = 1$.
• $n$-th powers → root test, using $\limsup a_n^{1/n}$. The root test is strictly stronger: whenever the ratio test decides, the root test decides the same way, and it sometimes decides when the ratio test cannot.
• Ratio gave exactly 1 → Raabe. Compute $n(a_n/a_{n+1} - 1) \to R$: converges for $R > 1$, diverges for $R < 1$.
• Logarithms, with terms decreasing → Cauchy condensation: $\sum a_n$ behaves like $\sum 2^k a_{2^k}$. This is what settles $\sum 1/(n \log n)$ (diverges) against $\sum 1/(n \log^2 n)$ (converges) — a pair the paper returns to often.

Alternating → Leibniz: if $a_n$ decreases to $0$ then $\sum (-1)^n a_n$ converges. Then check absolute convergence separately, because Leibniz says nothing about it.

Absolute versus conditional

A series converges absolutely if $\sum |a_n|$ converges. Absolute convergence is what buys you the right to rearrange terms, multiply series together, and regroup freely.

Conditional convergence — convergent but not absolutely — buys you none of that. Riemann's rearrangement theorem says a conditionally convergent series can be rearranged to sum to any real number you name, or to diverge. The terms are the same; only their order changed.

That is worth sitting with, because it is genuinely strange and it is examined directly.

Where intuition breaks

"$a_n \to 0$, so $\sum a_n$ converges." $\sum 1/n$. Everyone knows this and a surprising number still tick the option, because the statement appears embedded in a longer one where the false half is easy to miss.

"$\sum a_n$ converges, so $\sum a_n^2$ converges." False. Take $a_n = (-1)^n/\sqrt n$: the series converges by Leibniz, and squaring gives the harmonic series.

Why it is tempting: squaring makes terms smaller, so surely convergence survives. It does — for non-negative terms. Squaring also destroys the cancellation that was doing all the work, and cancellation is exactly what a conditionally convergent series depends on. This is a hypothesis dropped: the missing hypothesis is non-negativity.

"$\sum a_n^2$ converges, so $\sum a_n$ converges." Also false, in the other direction: $a_n = 1/n$.

"$\sum a_n$ and $\sum b_n$ converge, so $\sum a_n b_n$ converges." Take $a_n = b_n = (-1)^n/\sqrt n$. Both converge; the product series is $\sum 1/n$.

Notice that $(-1)^n/\sqrt n$ has now answered three separate false claims. It is the universal spoiler here, and it is worth carrying into the hall for that reason alone.

"The ratio test was inconclusive, so the root test will be too." False, and this is the one that catches strong candidates. Take $a_n = 2^{-n + (-1)^n}$. The ratios alternate between $2$ and $1/8$ and have no limit at all. But $a_n^{1/n} \to 1/2$, so the root test proves convergence outright. A converse assumed, dressed as a fact about the tests rather than about a series.

"A convergent series can be rearranged freely." Only if it converges absolutely. Riemann again — and options phrased as "the sum is unchanged under rearrangement" are testing whether you noticed the word absolutely was missing.

"$\sum a_n$ converges with $a_n \ge 0$ decreasing, so $n a_n \to 0$." This one is true — Pringsheim's theorem — and it appears among false statements precisely because it looks like the others. Do not reject an option merely because it feels like a trap.

The exam's angle

1. Check $a_n \to 0$ first. It is free, and it settles a fair number of options.
2. Positive or alternating? Alternating sends you to Leibniz, and then to a separate question about $\sum |a_n|$.
3. Match the shape to the test. Factorials → ratio. Powers of $n$ → root. Logs → condensation.
4. If the option mentions rearrangement, squares, or products, check for absolute convergence. Almost every false option in this family is a statement that would be true with absolute convergence and is false without it.

The night before

• $a_n \to 0$ is necessary, never sufficient. $\sum 1/n$.
• $\sum 1/n^p$ converges iff $p > 1$. $\sum 1/(n\log n)$ diverges; $\sum 1/(n \log^2 n)$ converges.
• Root is stronger than ratio. Ratio inconclusive does not mean root inconclusive: $2^{-n+(-1)^n}$.
• Raabe rescues ratio $= 1$; condensation handles logarithms.
• $(-1)^n/\sqrt n$ is the universal spoiler: kills squares, products, and "converges ⇒ converges absolutely".
• Absolute convergence is what permits rearranging, multiplying and regrouping. Riemann: a conditionally convergent series can be rearranged to any sum at all.
• Useful values: $\sum 1/n^2 = \pi^2/6$, $\sum (-1)^{n+1}/n = \ln 2$, $\sum (-1)^n/(2n+1) = \pi/4$.