Infinite Series and Convergence

1. Sequences versus series

The two objects are constantly confused, and the confusion costs more than it should. A sequence is an ordered list of numbers $a_1, a_2, a_3, \ldots$ — you can ask what they approach. A series is what happens when you put plus signs between them, $a_1 + a_2 + a_3 + \cdots$ — you can ask what that infinite addition adds up to. Those are different questions with different answers.

We met this distinction already in Algebra · Sequences & Series, where the geometric series gave us a first taste of what an infinite sum even means. Here we make it rigorous, then build the machinery — convergence tests — that lets you decide the question for series the geometric formula can't touch.

2. Partial sums and the definition of convergence

You can't actually add infinitely many things. What you can do is add the first $n$ of them and watch what happens as $n$ grows.

The partial sums form a brand-new sequence $S_1, S_2, S_3, \ldots$, and the whole "does the infinite sum exist" question reduces to a question about that sequence.

Two things to notice. First, "the sum of the series" is a definition, not an arithmetic operation — there's no infinite addition happening; there's a limit. Second, the whole apparatus of limits (§Limits) is doing the actual work. Convergence of series is convergence of sequences in disguise.

A first necessary condition

If $S_n \to S$, then consecutive partial sums get arbitrarily close, so $a_n = S_n - S_{n-1} \to 0$. Contrapositive:

3. The geometric series, revisited

You met the geometric series in algebra. Here it is, restated in the partial-sum language and with full force:

The partial sum has a closed form (the "shift-and-subtract" trick from algebra):

What happens to $r^{n+1}$ as $n \to \infty$ is the whole story. If $|r| < 1$, it goes to zero, and

If $|r| \geq 1$, the term $r^{n+1}$ either runs to infinity or oscillates without settling — the partial sums don't converge.

Telescoping series — the other one that just works

A series of the form $\sum (b_n - b_{n+1})$ collapses on summation: $S_n = b_1 - b_{n+1}$, so it converges to $b_1 - \lim b_n$ exactly when that limit exists. The classic example:

Geometric and telescoping series are the two patterns where you actually compute the sum. Everything else, you mostly settle for "converges" or "diverges" and stop there.

4. The harmonic series diverges

The harmonic series is

Its terms shrink to zero, so the $n$-th term test is silent. And yet it diverges. The cleanest argument is Nicole Oresme's grouping proof from around 1350.

Oresme's grouping proof

Group the terms after the first into blocks of size $1, 2, 4, 8, \ldots$:

Each block of $2^k$ terms is bounded below by replacing every entry with the smallest one, $1/2^{k+1}$, repeated $2^k$ times: that gives exactly $1/2$. There are infinitely many blocks, each contributing at least $\tfrac{1}{2}$. So the partial sums grow without bound — slowly (in fact $S_n \sim \ln n$), but inescapably.

5. The $p$-series

Generalising both the harmonic series and its "nice" cousins:

So $\sum 1/n^{1.5}$ converges, $\sum 1/n^2 = \pi^2/6$ converges (Euler's celebrated result), $\sum 1/n$ diverges (just barely), and $\sum 1/\sqrt n$ diverges too. The $p$-series is the second standard benchmark — together with the geometric series, it's the family you'll keep comparing other series to.

The proof — for $p > 1$ converges, $p \leq 1$ diverges — is one line each via the integral test, which is coming up next.

6. Convergence tests, in the order you reach for them

No single test handles every series. The skill is matching a test to a series's shape — and applying the cheap ones first. Here's the working list, roughly in the order of "try this before that one."

6.1 The $n$-th term test (always free)

Look at $a_n$. If it doesn't go to zero, the series diverges and you're done. If it does go to zero, you've learned nothing — go on.

6.2 Recognise the form

Before reaching for a real test, ask: is this a geometric series? A $p$-series? A telescoping series? Each has an immediate verdict. The fastest test is no test at all.

6.3 The integral test

Suppose $f$ is positive, continuous, and decreasing on $[1, \infty)$, and $a_n = f(n)$. Then

either both converge or both diverge. The picture is rectangle-versus-area: the sum is the total area of unit-width rectangles, and $f$ is monotone, so the rectangles and the integral sandwich each other within a constant.

Use it when $f(n)$ has a clean antiderivative — anything involving $1/n^p$, $1/(n \ln n)$, $e^{-n}$, and friends.

6.4 Direct comparison test

For series of non-negative terms $0 \leq a_n \leq b_n$:

• If $\sum b_n$ converges, so does $\sum a_n$ (you're below something finite).
• If $\sum a_n$ diverges, so does $\sum b_n$ (you're above something that runs away).

The direction matters and is easy to mix up. Memorise: convergence pushes down, divergence pushes up.

6.5 Limit comparison test

Often the bounds in direct comparison are awkward to set up — the algebra fights you. The limit comparison test sidesteps this. If $a_n, b_n > 0$ and

then $\sum a_n$ and $\sum b_n$ behave the same — both converge or both diverge. The standard play: pick $b_n$ to be the "leading-order behaviour" of $a_n$ for large $n$. For $a_n = \dfrac{2n+1}{n^3 + 3n}$, take $b_n = 1/n^2$, get $L = 2$, and inherit convergence from the $p$-series.

6.6 Ratio test

Compute $\rho = \lim_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n}\right|$.

• $\rho < 1$: $\sum a_n$ converges (absolutely).
• $\rho > 1$ (including $\infty$): $\sum a_n$ diverges.
• $\rho = 1$: inconclusive — try something else.

The ratio test eats factorials and exponentials for breakfast. Whenever you see $n!$, $r^n$, or anything whose successive terms relate by a clean ratio, this is the test.

6.7 Root test

Compute $\rho = \lim_{n \to \infty} \sqrt[n]{|a_n|}$.

• $\rho < 1$: converges absolutely.
• $\rho > 1$: diverges.
• $\rho = 1$: inconclusive.

Use it when $a_n$ is itself an $n$-th power — $\left(\tfrac{n}{n+1}\right)^{n^2}$, $(\ln n)^{-n}$, things like that. Ratio and root tests are siblings; when both apply, they agree, but each is awkward where the other shines.

6.8 Alternating series test (Leibniz)

For a series whose signs strictly alternate, $\sum (-1)^{n} b_n$ with $b_n > 0$:

If $b_n$ is monotonically decreasing and $b_n \to 0$, the series converges.

Reference table

8. Absolute vs. conditional convergence

Once the terms of a series can take both signs, "converges" splits in two.

A foundational theorem ties them together:

If $\sum |a_n|$ converges, then $\sum a_n$ converges. Absolute convergence implies convergence.

The strategy that falls out of this: to check whether a signed series converges, run any positive-term test (comparison, ratio, root, integral) on $\sum |a_n|$. If it converges, you're done — you have absolute convergence, which implies convergence. Only if $\sum |a_n|$ diverges do you have to be subtle.

The alternating harmonic series

The cleanest example of the two regimes living right next to each other:

By the alternating series test, $b_n = 1/n$ is positive, decreasing, and tends to zero, so the series converges. But $\sum |a_n| = \sum 1/n$ is the harmonic series, which diverges. So this series converges conditionally — its convergence is a delicate cancellation of positives against negatives.

Why the distinction matters: rearrangement

For absolutely convergent series, by contrast, every rearrangement gives the same sum. This is why "absolute convergence" is the version of convergence that lets you safely manipulate series — regroup them, multiply them, integrate or differentiate them term-by-term. Conditional convergence is real, but fragile. The rule of thumb: if you're going to do anything beyond reading the series, you want absolute convergence.

9. Worked examples for each test

One example per test, with the reasoning written out. Try each before opening the solution and see whether your matching instinct lines up with the one used.

10. Common pitfalls