Series & Convergence

Infinite sums, tests for convergence, power series, and Taylor series โ€” when adding infinitely many terms gives a finite and useful answer.

Partial sums of geometric series ฮฃ rแต โ€” converges to 1/(1โˆ’r) when |r| < 1
L = 2.0000124567810111200.460.921.381.842.3Partial sum = 1.9995
r = 0.500.95
n = 1230
Definition

An infinite series is the sum of infinitely many terms: โˆ‘n=1โˆžan=a1+a2+a3+โ‹ฏ\displaystyle\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \cdots

To define this rigorously, form the partial sums SN=โˆ‘n=1NanS_N = \sum_{n=1}^N a_n. The series converges to LL if SNโ†’LS_N \to L as Nโ†’โˆžN \to \infty; otherwise it diverges.

Geometric series: โˆ‘n=0โˆžrn=11โˆ’r\displaystyle\sum_{n=0}^\infty r^n = \frac{1}{1-r} for โˆฃrโˆฃ<1|r| < 1; diverges for โˆฃrโˆฃโ‰ฅ1|r| \geq 1.

The nn-th term test: if limโกnโ†’โˆžanโ‰ 0\displaystyle\lim_{n\to\infty} a_n \neq 0, the series diverges. (The converse is false: anโ†’0a_n \to 0 does not guarantee convergence.)

Geometric series

12+14+18+โ‹ฏ=โˆ‘n=1โˆž(12)n=1/21โˆ’1/2=1\displaystyle\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n = \frac{1/2}{1 - 1/2} = 1

This resolves Zeno's paradox: infinitely many positive terms can sum to a finite value.

Try it

Determine whether โˆ‘n=1โˆžn2n+1\displaystyle\sum_{n=1}^\infty \frac{n}{2n+1} converges.

Solution

limโกnโ†’โˆžn2n+1=12โ‰ 0\displaystyle\lim_{n\to\infty} \frac{n}{2n+1} = \frac{1}{2} \neq 0.

By the nn-th term test, the series diverges.

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