# Formula to estimate sum to nearly correct : $\sum_{n=0}^\infty\frac{(-1)^n}{n^3}$

Estimate the sum correct to three decimal places : $$\sum_{n=0}^\infty\frac{(-1)^n}{n^3}$$

This problem is in my homework. I find that n = 22 when use Maple to solve this. (with some programming) But, in my homework, teacher said that find formula for this problem.

Thanks :)

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Does that sum really start at $n=0$? –  WimC Jun 10 '12 at 17:08

By the Alternating Series Test, the error to an alternating series with monotonically decreasing terms is the next term to be added. Thus, to get three decimal places, we would need to find an $n$ so that $n^3>2000$, which would be $n=13$. Thus, summing the first 12 terms should get you to within 3 decimal places.

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To get 3 decimal places accuracy, don't we need $n^3> 10,000$? (And the cube root of 10,000 is indeed about 21.5.) –  Ragib Zaman Jun 10 '12 at 15:53
Three decimal places would mean an error of less than $.0005=\frac{1}{2000}$. –  robjohn Jun 10 '12 at 15:58
In practice here even 22 terms is not enough to get the correct result (rounded!) since $s(22)=-0.90149891\cdots$ and $s(23)=-0.90158110\cdots$ (23 is fine since $s(24)=-0.901508\cdots$ !). –  Raymond Manzoni Jun 10 '12 at 16:02
@RaymondManzoni: That is true. To make sure the first $3$ decimal places are correct, it could require many more places to be computed to make sure that rounding and subsequent terms don't affect the first three digits. That is why I consider "to $3$ places" to mean "an error of less than $.0005$" (and this is simply my opinion). –  robjohn Jun 10 '12 at 16:13
@robjohn: I understand your point of view (mine would be similar) as well as Ragib's. It was just fun to notice that finding the number of terms required was more prickly than it seemed... –  Raymond Manzoni Jun 10 '12 at 16:22
For alternating sums $\sum(-1)^n a_n$ with $a_n> 0$ strictly decreasing there is a simple means to estimate the remainder $\sum^\infty_{k=n} (-1)^k a_k$. You can just use $a_{n-1}$.