# Remainder: Alternate Series

1. Oct 1, 2006

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

2. Oct 1, 2006

### Hurkyl

Staff Emeritus
Well, the easy answer is if |b_n| is decreasing, then the sum of the entire series lies between any two consecutive partial sums.

3. Oct 1, 2006

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?