AP Calculus BC glossary

Convergent Series

Also called: Series convergence

A convergent series is one whose partial sums settle on a finite limit, and that limit is its sum. Where the partial sum has a closed form, as it does for telescoping and geometric series, the sum comes straight out of it. Dropping finitely many terms changes the sum but never changes convergence.

∑n=1∞an=L  ⟺  lim⁡N→∞SN=L∈R\sum_{n=1}^{\infty} a_n = L \iff \lim_{N \to \infty} S_N = L \in \mathbb{R}

Convergence is settled by the partial sums, and sometimes you can write one down in closed form instead of testing it. For ∑1/(n(n+1))\sum 1/(n(n+1)), splitting each term as 1/n−1/(n+1)1/n - 1/(n+1) makes the inside of SNS_N cancel and leaves two survivors.

SN=∑n=1N(1n−1n+1)=1−1N+1⟶1S_N = \sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+1}\right) = 1 - \frac{1}{N+1} \longrightarrow 1

Convergence never depends on the front of a series. Change or delete the first thousand terms and the sum shifts by a finite amount, while convergence itself is untouched, because only the tail has to settle. That is why every test is stated for nn large enough, and why a comparison only has to hold eventually.

The mistake

Cancelling the middle of a telescoping sum and stopping there. In ∑n=1∞(1n−1n+2)\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+2}\right) two positive terms survive at the front and two negative ones at the back, so SN=32−1N+1−1N+2S_N = \frac{3}{2} - \frac{1}{N+1} - \frac{1}{N+2} and the sum is 32\frac{3}{2}, not 11. Write SNS_N out before you take the limit.

Appears in: Unit 10: Infinite Sequences and Series (BC)