Turning an infinite sum into a sequence
Writing a_1 + a_2 + a_3 + … looks like it asks us to perform infinitely many additions, which no one can do. The trick of analysis is to never add infinitely many numbers at once. Instead, from the terms a_n we build a brand-new sequence, the sequence of partial sums: s_1 = a_1, s_2 = a_1 + a_2, s_3 = a_1 + a_2 + a_3, and in general s_N = a_1 + … + a_N. Each s_N is an honest finite sum.
Now we define the infinite series sum a_n to converge to a number s precisely when the sequence of partial sums has limit s, that is s_N -> s. In that case we call s the sum and write sum a_n = s. If the partial sums have no finite limit, the series diverges. So a convergent series is nothing more — and nothing less — than a convergent sequence wearing a disguise.
The geometric series, worked in full
The one series whose sum we can compute by hand is the geometric series sum_{n=0}^∞ r^n. Its partial sums collapse to a closed form, and from there the limit is immediate. This is worth memorizing because almost every later test secretly compares against it.
Claim: for |r| < 1, sum_{n=0}^∞ r^n = 1/(1 - r).
Step 1. Write the N-th partial sum and multiply by r.
s_N = 1 + r + r^2 + ... + r^N
r*s_N = r + r^2 + ... + r^N + r^(N+1)
Step 2. Subtract to telescope the middle.
s_N - r*s_N = 1 - r^(N+1)
(1 - r) s_N = 1 - r^(N+1)
s_N = (1 - r^(N+1)) / (1 - r) [valid since r != 1]
Step 3. Take the limit N -> infinity.
If |r| < 1 then r^(N+1) -> 0, so
s_N -> 1/(1 - r).
Therefore the series converges to 1/(1 - r).
Step 4. If |r| >= 1 the terms r^n do NOT tend to 0,
so s_N has no finite limit and the series diverges.
Example: r = 1/2 gives 1 + 1/2 + 1/4 + ... = 1/(1 - 1/2) = 2.Notice exactly where convergence is decided: in Step 3, by whether r^(N+1) -> 0. The infinite sum is real only because that one sequence limit exists. Everything rides on the partial sums.
Telescoping: when the sum collapses
A second family we can sum exactly is the telescoping series, where each term is a difference b_n - b_{n+1}. Almost everything cancels, leaving only the ends. It is the geometric trick's cousin and a clean place to see the partial-sum philosophy in action.
Sum sum_{n=1}^∞ 1/(n(n+1)).
Partial fractions: 1/(n(n+1)) = 1/n - 1/(n+1).
s_N = sum_{n=1}^N (1/n - 1/(n+1))
= (1 - 1/2) + (1/2 - 1/3) + ... + (1/N - 1/(N+1))
= 1 - 1/(N+1) [every interior term cancels]
Limit: s_N = 1 - 1/(N+1) -> 1 as N -> infinity.
Therefore sum_{n=1}^∞ 1/(n(n+1)) = 1.