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10.1 Defining Convergent and Divergent Infinite Series

Syllabus
2020
Topic
10.1
Level
—

Learning objectives

—LIM-7.A—Determine whether a series converges or diverges• LIM-7.A Determine whether a series converges or diverges.• LIM-7.A.1 The nth partial sum is defined as the sum of the first n terms of a series.• LIM-7.A.2 An infinite series of numbers converges to a real number S (or has sum S), if and only if the limit of its sequence of partial sums exists and equals S.• Enduring understanding LIM-7: Applying limits may allow us to determine the finite sum of infinitely many terms.

A Series Converges When Its Partial Sums Settle

For the infinite series ∑k=1∞ak\sum_{k=1}^{\infty}a_k, the nnth partial sum SnS_n adds only the first nn terms. The infinite series converges to SS exactly when the sequence S1,S2,S3,…S_1,S_2,S_3,\ldots approaches the finite real number SS.

S_n=\sum_{k=1}^{n}a_k,\qquad \sum_{k=1}^{\infty}a_k=S\iff\lim_{n\to\infty}S_n=S

  1. Write a formula for the finite partial sum SnS_n.
  2. Simplify SnS_n before taking a limit.
  3. Evaluate lim⁡n→∞Sn\lim_{n\to\infty}S_n.
  4. If the limit is a finite real number, the series converges to it; if no finite limit exists, the series diverges.

Example: since 1/[k(k+1)]=1/k−1/(k+1)1/[k(k+1)]=1/k-1/(k+1),
Sn=∑k=1n1k(k+1)=(1−12)+(12−13)+⋯+(1n−1n+1)=1−1n+1.S_n=\sum_{k=1}^{n}\frac{1}{k(k+1)}=\left(1-\frac12\right)+\left(\frac12-\frac13\right)+\cdots+\left(\frac1n-\frac1{n+1}\right)=1-\frac1{n+1}.
Therefore lim⁡n→∞Sn=1\lim_{n\to\infty}S_n=1, so ∑k=1∞1/[k(k+1)]\sum_{k=1}^{\infty}1/[k(k+1)] converges and its sum is 11. By contrast, for ∑k=1∞1\sum_{k=1}^{\infty}1, Sn=n→∞S_n=n\to\infty, so the series diverges.

A convergent series still contains infinitely many terms; convergence does not mean the adding stops. It means the finite partial sums approach one finite value. Writing that a divergent series has a sum of ∞\infty is not the same as convergence to a real number.