<s>[INST] Formula to estimate sum to nearly correct : $\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$ Estimate the sum correct to three decimal places : 
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^3}$$
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 find the formula for this problem.
Thanks :)
 [/INST] Averaging the 9th and 10th partial sums will do it, as in robjohn's answer and my comment there. Just for fun, as an alternative to alternating series methods, you could group consecutive terms to give a positive series and then use integral approximation.
$$\begin{align}
S:=\sum_{n=1}^\infty\frac{(-1)^n}{n^3}
&=-1+\sum_{n=1}^\infty\frac{1}{8n^3}-\frac{1}{(2n+1)^3}=-1+\sum_{n=1}^\infty f(n)\\
\end{align}
$$
And then
$$\begin{align}
-1+\sum_{n=1}^{N}\frac{1}{8n^3}-\frac{1}{(2n+1)^3}+\int_{N}^\infty\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx<S<-1+\sum_{n=1}^N\frac{1}{8n^3}-\frac{1}{(2n+1)^3}+\int_{N-1}^\infty\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx\\
\end{align}
$$
The difference between the outer bounds is 
$$\begin{align}
\int_{N-1}^{N}\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx
&=\left[{-{\frac1{16x^2}}}+\frac{1}{4(2x+1)^2}\right]_{N-1}^{N}\\
&={-{\frac1{16N^2}}}+\frac{1}{4(2N+1)^2}+{{\frac1{16(N-1)^2}}}-\frac{1}{4(2N-1)^2}\\
\end{align}$$
We'd like this difference to be less than $0.001$. This way we could average the two outer bounds as an estimate for the sum and know that our estimate was within $0.0005$. This happens for $N=5$. So an estimate that will work out to within $0.0005$ is 
$$\sum_{n=1}^{11}\frac{(-1)^n}{n^3}+\frac12\int_4^{5}\frac{1}{8x^3}-\frac{1}{(2x+1)^3}\,dx$$
which works out to 
$$\sum_{n=1}^{11}\frac{(-1)^n}{n^3}+\frac{24209}{125452800}\approx-0.9016748\ldots$$
</s>