Calculus I · Adding Up Forever

The one sum you can compute exactly: the geometric series

Almost no series can be summed in closed form. One family can, and it’s the benchmark for everything else — and you’ve already met it, as the Taylor series of $\tfrac{1}{1-x}$ in Chapter 7. The geometric series adds powers of a fixed ratio $r$:

$$\sum_{n=0}^\infty r^n = 1 + r + r^2 + r^3 + \cdots.$$

Its partial sum has a clean form (multiply $S_N = 1 + r + \cdots + r^N$ by $r$ and subtract): $S_N = \dfrac{1 - r^{N+1}}{1 - r}$. Now take $N \to \infty$. If $|r| < 1$, then $r^{N+1} \to 0$ and the sum converges:

$$\boxed{\;\sum_{n=0}^\infty r^n = \frac{1}{1-r} \quad (|r| < 1).\;}$$

If $|r| \ge 1$, the terms don’t shrink and the partial sums run away — divergent. So $\tfrac12 + \tfrac14 + \cdots = \tfrac{1}{1-1/2} = 1$ ✓, while $1 + 2 + 4 + \cdots$ explodes. This is precisely Chapter 7’s $\tfrac{1}{1-x} = 1 + x + x^2 + \cdots$ with $x = r$, and the condition $|r|<1$ is the same “wall at $\pm 1$” you saw the series slam into there. The geometric series is the ruler we measure other series against.

A first sanity check: the terms must die

Here is the simplest possible test, and the one principle you can always apply first.

The term test. If $\displaystyle\sum a_n$ converges, then $a_n \to 0$.

The reason is one line: $a_n = S_n - S_{n-1}$, and if both partial sums converge to the same $S$, their difference goes to $S - S = 0$. The useful form is the contrapositive: if the terms do not go to zero, the series diverges. So $\sum \tfrac{n}{n+1}$ diverges instantly (its terms approach $1$, not $0$) — no further work needed.

But beware the trap that ensnares everyone at first: the term test is necessary, not sufficient. Terms going to zero does not guarantee convergence. The most important example in the subject proves it.

The harmonic series: every term vanishes, yet the sum is infinite

Consider $\displaystyle\sum_{n=1}^\infty \frac1n = 1 + \frac12 + \frac13 + \frac14 + \cdots$. The terms $\tfrac1n$ shrink to zero, so the term test is passed — and yet this series diverges to infinity. This is one of the genuine surprises of mathematics, and a medieval one: the argument is due to Nicole Oresme, around 1350, and it is pure cleverness.

◆ Layer 2 — Oresme’s proof that the harmonic series diverges

Group the terms in blocks of doubling length and bound each block from below: $$1 + \frac12 + \underbrace{\left(\frac13 + \frac14\right)}_{> \,\frac14 + \frac14 = \frac12} + \underbrace{\left(\frac15 + \frac16 + \frac17 + \frac18\right)}_{>\, 4\cdot\frac18 = \frac12} + \underbrace{\left(\frac19 + \cdots + \frac1{16}\right)}_{>\, 8\cdot\frac1{16} = \frac12} + \cdots$$ Every block exceeds $\tfrac12$ (each has terms no smaller than its last, and there are exactly enough of them). Since there are infinitely many blocks, each contributing more than $\tfrac12$, the partial sums climb past $\tfrac12, 1, \tfrac32, 2, \ldots$ without bound. The sum is infinite. $\blacksquare$

Keep the harmonic series close; it is the dividing line. “The terms go to zero” feels like it should be enough — and the harmonic series is the eternal reminder that it isn’t. To decide convergence, we need real tools. Here are the two.

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