When Infinite Sums Have Finite Answers

Add together infinitely many numbers. What do you get?

Sometimes infinity. Sometimes a perfectly ordinary finite number. The difference depends on how quickly the terms shrink — and this turns out to be one of the subtler, more surprising corners of mathematics.

The geometric series: the cleanest example

If you multiply a starting value by the same ratio repeatedly and add everything up:

$S = 1 + r + r^2 + r^3 + \cdots$

There's an elegant trick. Multiply $S$ by $r$: $rS = r + r^2 + r^3 + \cdots$

Subtract: $S - rS = 1$, so $S(1-r) = 1$, giving $S = \frac{1}{1-r}$.

This works when $|r| < 1$ — when the ratio is small enough that the terms shrink to zero fast enough.

The harmonic series: deceptively divergent

Now consider $1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots$

Each term is smaller than the last. Surely this converges?

It doesn't. The harmonic series diverges — the partial sums grow without bound.

The proof is one of the most elegant in mathematics, due to Nicole Oresme around 1350:

$1 + \frac{1}{2} + \underbrace{\frac{1}{3} + \frac{1}{4}}_{\geq \frac{1}{4}+\frac{1}{4}=\frac{1}{2}} + \underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}}_{\geq 4 \cdot \frac{1}{8}=\frac{1}{2}} + \cdots$

Every block of terms (doubling in size) contributes at least $\frac{1}{2}$. Since you can form infinitely many blocks, the sum is infinite.

The terms go to zero — but not fast enough. This is the key lesson: $a_n \to 0$ is necessary for convergence, but not sufficient.

How slowly does the harmonic series diverge?

Almost comically slowly. The partial sum $H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n}$ grows like $\ln n$.

• To reach $H_n = 10$: you need about $n = e^{10} \approx 22{,}000$ terms.
• To reach $H_n = 100$: about $n \approx 10^{43}$ terms.

No computer could ever compute enough terms to reach $H_n = 100$ by direct summation.

The Basel problem: a beautiful surprise

Now add the squares of the reciprocals:

$1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \cdots = \frac{\pi^2}{6}$

This is the Basel problem, solved by Leonhard Euler in 1735. The result shocked the mathematical world. Why does $\pi$ — a number about circles — appear in the sum of reciprocals of squares?

Euler's proof used the fact that $\sin x / x$ can be expressed as an infinite product over its roots, which he compared to a polynomial's factored form. The argument is not rigorous by modern standards, but the answer is correct, and a rigorous proof was found within decades.

The appearance of $\pi$ is not coincidence — it reflects a deep connection between number theory and analysis that runs through the Riemann zeta function.

Convergence tests in brief

How do you tell whether a series converges?

Ratio test: compute $L = \lim |a_{n+1}/a_n|$. If $L < 1$, converges. If $L > 1$, diverges.

Comparison: if your terms are smaller than a convergent series' terms, you converge.

$p$-series: $\sum \frac{1}{n^p}$ converges iff $p > 1$. Borderline case at $p = 1$ (harmonic, diverges).

Alternating series test: alternating signs with decreasing terms going to zero — converges.

Absolute versus conditional convergence

Some series converge because of cancellation between positive and negative terms, not because the terms shrink fast enough.

The alternating harmonic series: $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots = \ln 2 \approx 0.693$

This converges. But $\sum \frac{1}{n}$ (ignoring signs) diverges. Such a series is called conditionally convergent.

The Riemann rearrangement theorem: if you rearrange the terms of a conditionally convergent series, you can make it converge to any value you like — or diverge to infinity, or oscillate forever.

This is genuinely strange. Addition is supposed to be commutative. But for conditionally convergent series, the order of the terms determines the sum.