# Remainder: Alternate Series

How many terms of the series do we need to add in order to find the sum to the indicated accuracy?

$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{2}}$$, $$| error | < 0.01$$. So, $$b_{n} = \frac{1}{n^{2}}$$. $$b_{n} < b_{n+1}$$, and $$\lim_{n\rightarrow \infty} b_{n} = 0$$. Therefore, the series is convergent. I wrote out some terms of the series: $$s = 1-\frac{1}{4}+\frac{1}{9}-\frac{1}{16}+\frac{1}{25}-\frac{1}{36}+\frac{1}{49}-\frac{1}{64}+\frac{1}{81}-\frac{1}{100}+\frac{1}{121}+ . . .$$. From this step, how do we determine the number of terms we need to add in order to find the sum to the indicated accuracy?

Thanks

Hurkyl
Staff Emeritus
oh ok, so just look at the difference between a $$b_{n+1}-b_{n}$$ and see if it is less than $$0.01$$? And $$n$$ is the number of terms you need?