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 :)

$$\sum_{n=1}^{\infty} \dfrac{(-1)^n}{n^3} = \sum_{k=1}^{\infty} \left(\dfrac{1}{k^3}-\dfrac{1}{(k+1)^3}\right)= \sum_{n=1}^{\infty} \left(\dfrac{(k+1)^3-k^3}{k^3(k+1)^3}\right)$$
$$= \sum_{k=1}^{\infty} \dfrac{(k+1-k)(k^2+k+1)}{k^3(k+1)^3}$$
$$= \sum_{k=1}^{\infty} \dfrac{k^2+k+1}{k^3(k+1)^3}$$
$ $
Note that for the sum to be accurate within 3 decimal planes the nth term must be less than 0.001
Therefore we have $$\dfrac{k^2+k+1}{k^3(k+1)^3} < \dfrac{1}{1000}$$
You will notice that answer can either be $2k$ or $2k+1$
I will think about it later, how to pin point it, I will have to go now, let me know what you guys think about it
Using wolfram you will find that $k=5.18$ satisfies the inequality
Therefore the smallest value of $k$ to satisfy the inequality will be $6$
And hence the answer can either be $12$ or $13$ 
In fact it should be $n=13$ terms, do you need me to explain that? Try thinking first, there is a clear logical reason for $n = 13$ and not $n=12$.
