The most important series in all of mathematics is the geometric series:
an
infinite sum
whose terms form a
geometric sequence,
each a fixed ratio r times the last,
\sum_{n=0}^{\infty} a r^{n} = a + ar + ar^2 + ar^3 + \cdots
with first term a \ne 0 and common ratio
r. Whether this converges turns entirely on the size of
r: if |r| < 1 each term is a fraction
of the last and the total settles; if |r| \ge 1 the terms do not
shrink and the sum runs away. And when it does converge, the value is a single clean formula.
The centerpiece: closed form, then the limit
We derive everything from scratch. First a formula for the
partial sum
s_N, then its limit.
Step 1 — Write the partial sum. Add the first
N+1 terms (indices 0 through
N):
s_N = a + ar + ar^2 + \cdots + ar^{N}.
Step 2 — Multiply the whole thing by r. Every
term's power steps up by one:
r\,s_N = ar + ar^2 + ar^3 + \cdots + ar^{N+1}.
Step 3 — Subtract. Line up s_N - r\,s_N: every
middle term ar, ar^2, \dots, ar^N appears in both lists
and cancels, leaving only the first term of s_N and the last of
r s_N:
s_N - r\,s_N = a - ar^{N+1}.
Step 4 — Factor and solve for s_N. The left side
is s_N(1 - r); the right is a(1 - r^{N+1}).
Provided r \ne 1 we may divide:
s_N = a\,\frac{1 - r^{\,N+1}}{1 - r}.
That is the closed form for the partial sum — exact, with no infinity in sight yet.
Step 5 — Take N \to \infty. The only place
N appears is the term r^{N+1}. Its fate
depends on |r|:
|r| < 1 \implies r^{\,N+1} \to 0, \qquad |r| \ge 1 \implies r^{\,N+1} \not\to 0.
Step 6 — Read off the sum when |r| < 1. The
r^{N+1} term vanishes, and the partial sums converge:
\sum_{n=0}^{\infty} a r^{n} = \lim_{N\to\infty} a\,\frac{1 - r^{\,N+1}}{1 - r} = a\,\frac{1 - 0}{1 - r} = \frac{a}{1 - r}. \qquad \blacksquare
For |r| \ge 1 the term r^{N+1} either
blows up (|r| > 1) or refuses to settle
(r = -1 oscillates; r = 1 gives
s_N = a(N+1) \to \pm\infty) — so the series diverges. The dividing
line is exactly |r| = 1.
For the geometric series \sum_{n=0}^{\infty} a r^{n} with a \ne 0:
-
it converges if and only if |r| < 1;
-
and then its sum is
\displaystyle \sum_{n=0}^{\infty} a r^{n} = \frac{a}{1 - r};
-
for |r| \ge 1 the terms do not shrink to
0, so the series diverges;
-
the partial sum (any r \ne 1) is
\displaystyle s_N = a\,\frac{1 - r^{N+1}}{1 - r}.
The surprising identity 0.999\ldots = 1 is simply a geometric
series in disguise. Read the repeating 9s as a sum:
0.999\ldots = \frac{9}{10} + \frac{9}{100} + \frac{9}{1000} + \cdots = \sum_{n=0}^{\infty} \frac{9}{10}\left(\frac{1}{10}\right)^{n},
a geometric series with a = \tfrac{9}{10} and
r = \tfrac{1}{10} (so |r| < 1). The
formula gives
\sum_{n=0}^{\infty} \frac{9}{10}\left(\frac{1}{10}\right)^{n} = \frac{a}{1-r} = \frac{9/10}{1 - 1/10} = \frac{9/10}{9/10} = 1.
Not "approximately" 1, not "infinitely close" —
exactly 1. The same machine turns any repeating
decimal into a fraction. For 0.\overline{27} = 0.272727\ldots,
take a = \tfrac{27}{100},
r = \tfrac{1}{100}:
0.\overline{27} = \frac{27/100}{1 - 1/100} = \frac{27/100}{99/100} = \frac{27}{99} = \frac{3}{11}.
Every repeating decimal is rational precisely because it is a convergent geometric series.