Example 1: ln

Approximate ln (1½) within 0.01. We use the power series for ln (1 - x),

, r = 1.

Setting 1 - x = 1½, x = -½,

This is an alternating series. The last term shown is less than 0.01,

1/(5 · 32) = 1/160 ~ 0.006.

By the Alternating Series Test, the error in each partial sum is less than the next term. So

, error ≤

or

ln (1½) ~ 0.401, error ≤ 0.006.

The actual value is

ln (1½) ~ 6.405.