Infinite series

An infinite series $a_1 + a_2 + a_3 + \cdots$ has no last term, so we look at the sum of its first n terms, the partial sum $S_n$, as n grows. If $S_n$ approaches a number S, the series converges and S is its sum. Otherwise it diverges.

The graph plots $S_n$ against n on a logarithmic axis from 1 to a million. The number line on the left draws each term as an arc from $S_{k-1}$ to $S_k$. Positive terms make arcs that all go one way, and alternating terms make arcs that swing back and forth.

Terms that tend to 0 are necessary for convergence, because $a_n = S_n - S_{n-1} \to S - S = 0$. They are not enough. The harmonic series $1 + \tfrac12 + \tfrac13 + \cdots$ has terms that tend to 0, yet its partial sums grow without bound. Integral comparison shows why: the rectangle of height 1/k on [k, k + 1] lies above the curve y = 1/x, so $S_n > \ln(n+1)$. The growth is so slow that a million terms add up to only about 14.39.

The same rectangles settle the p-series $\sum 1/k^p$: it converges exactly when p > 1. For p > 1 they also pin down what is left after n terms, the remainder $R_n = S - S_n$:

So the page can give an interval that must contain the limit, even when the limit has no simple formula. A narrow interval does not mean that $S_n$ is close, though. At p = 1.1, a million terms still fall about 2.5 short of the limit. Yet adding $S_n$ to the two integrals gives a tiny interval that contains it.

The geometric series $r + r^2 + r^3 + \cdots$ has partial sums with a closed form:

With r = ½ each term fills half of the gap still open, which is what Remaining gap draws, and $\tfrac12 + \tfrac14 + \tfrac18 + \cdots = 1$. At r = 1 the partial sums grow without bound, and at r = −1 they jump between −1 and 0 forever.

If the terms alternate in sign and their sizes decrease to 0, the series converges, and its sum always lies between two neighbouring partial sums. Each such interval sits inside the one before, and their widths shrink to 0, so they close down on a single number. Nested intervals stacks them. The page has two such series: $1 - \tfrac12 + \tfrac13 - \cdots = \ln 2$, and Leibniz’s $1 - \tfrac13 + \tfrac15 - \cdots = \pi/4$.

The alternating harmonic series converges only conditionally: without the signs, $\sum 1/k$ diverges. Riemann’s rearrangement theorem says that such a series can be reordered to converge to any number T. Rearranged keeps the terms of each sign in their original order: at or below T it adds the next positive term, above T the next negative one. The positive terms alone sum to +∞ and the negative terms to −∞, so each run reaches or crosses T. Each overshoot is at most the size of the term that caused it, and the terms shrink to 0, so the sums settle on T.