What could it mean to add up infinitely many numbers? You cannot perform infinitely
many additions. The trick is to add finitely many, watch where the running total goes, and
define the infinite sum as that destination — a
limit of a sequence.
Given a sequence of terms a_1, a_2, a_3, \dots, build the
partial sums — the totals after 1, 2, 3, \dots
terms:
s_N = a_1 + a_2 + \dots + a_N = \sum_{n=1}^{N} a_n.
The infinite series \sum_{n=1}^{\infty} a_n is
defined to be the limit of these partial sums, when that limit exists:
\sum_{n=1}^{\infty} a_n \;:=\; \lim_{N \to \infty} s_N.
If (s_N) converges to a number S, the
series converges and we write
\sum a_n = S. If (s_N) diverges, so
does the series — the infinite sum has no value. The whole subject of series is the study of
one sequence: the sequence of partial sums.
A telescoping series, line by line
The cleanest convergent series to evaluate by hand is the telescoping one,
where consecutive terms cancel and the partial sum collapses to almost nothing. We compute
\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} + \cdots
Step 1 — Split each term by partial fractions. Look for constants with
\tfrac{1}{n(n+1)} = \tfrac{A}{n} + \tfrac{B}{n+1}. Clearing
denominators gives 1 = A(n+1) + Bn; setting
n = 0 yields A = 1, and
n = -1 yields B = -1. So
\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}.
Step 2 — Write out the partial sum with the split terms.
s_N = \sum_{n=1}^{N}\left(\frac{1}{n} - \frac{1}{n+1}\right) = \left(\frac11 - \frac12\right) + \left(\frac12 - \frac13\right) + \cdots + \left(\frac1N - \frac1{N+1}\right).
Step 3 — Watch the middle collapse (the telescope). Every
-\tfrac1k is cancelled by the +\tfrac1k
in the next bracket. All that survives is the very first term and the very last:
s_N = \frac11 - \frac{1}{N+1} = 1 - \frac{1}{N+1}.
Step 4 — Take the limit of the partial sums. As
N \to \infty, the leftover tail
\tfrac{1}{N+1} \to 0 (it is the harmonic sequence shifted by one),
so
\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \lim_{N\to\infty} s_N = \lim_{N\to\infty}\left(1 - \frac{1}{N+1}\right) = 1. \qquad \blacksquare
Infinitely many positive terms, and yet they total exactly 1 —
because each new term is small enough that the running total is reined in.
For a sequence of terms (a_n), define the partial sums
s_N = \sum_{n=1}^{N} a_n. Then the series
\sum_{n=1}^{\infty} a_n is defined by
\sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} s_N,
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and the series converges exactly when the sequence
(s_N) of partial sums converges.
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The divergence test: if
\sum a_n converges then a_n \to 0.
Contrapositive — if the terms do not shrink to 0, the
series diverges. (The converse fails: see the harmonic series below.)
The divergence test is one-directional, and the harmonic series
\sum_{n=1}^{\infty} \tfrac1n is the reason why. Its terms
\tfrac1n \to 0 shrink to zero — yet the series
diverges to +\infty. Here is Oresme's grouping
argument (c. 1350). Bundle the terms in blocks whose sizes double:
1 + \underbrace{\frac12}_{} + \underbrace{\frac13 + \frac14}_{2\text{ terms}} + \underbrace{\frac15 + \frac16 + \frac17 + \frac18}_{4\text{ terms}} + \cdots
Replace every term in a block by the smallest one in it (the last). Then
\tfrac13 + \tfrac14 > \tfrac14 + \tfrac14 = \tfrac12, and
\tfrac15 + \dots + \tfrac18 > 4\cdot\tfrac18 = \tfrac12, and so
on — every block exceeds \tfrac12:
\sum_{n=1}^{\infty} \frac1n > 1 + \frac12 + \frac12 + \frac12 + \cdots = \infty.
Adding \tfrac12 infinitely often is unbounded, so the partial
sums climb past every ceiling. Terms shrinking to zero is necessary for
convergence but never sufficient — the harmonic series is the eternal cautionary
tale.