<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] With a little bit of approximation, we can achieve the desired $5\times 10^{-4}$ accuracy with fewer terms.
Notice
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^3} = -\left[\sum_{n=1}^{\infty}\frac{1}{n^3}- \sum_{n=1}^\infty\frac{2}{(2n)^3}\right] = -\frac34\sum_{n=1}^\infty \frac{1}{n^3} = -\frac34\zeta(3)$$
If we want to estimate the sum at the left accurate up to $5\times 10^{-4}$, 
it just suffices to estimate $\zeta(3)$ accurate up to $6.67\times 10^{-4}$. 
In the sum of $\zeta(3)$, if we pick a $m$ and replace $\displaystyle\;\frac{1}{n^3}$ by $\displaystyle\;\frac{1}{n^3-n}$ for $n \ge m$, the error introduced $\mathcal{E}_m$ is
$$\mathcal{E}_m =_{def} \sum_{n=m}^\infty\left(\frac{1}{n^3-n} - \frac{1}{n^3}\right)
= \sum_{n=m}^\infty \frac{1}{(n^2-1)n^3} \le \frac{m^2}{m^2-1}\sum_{n=m}^\infty \frac{1}{n^5}
$$
Since $\displaystyle\;\frac{1}{x^5}$ is a convex function, we have
$$\frac{1}{n^5} \le \int_{n-1/2}^{n+1/2} \frac{dx}{x^5}
\quad\implies\quad \mathcal{E}_m \le \frac{m^2}{m^2-1}\int_{m-1/2}^\infty \frac{dx}{x^5}
= \frac{m^2}{4(m^2-1)(m-1/2)^4}
$$
For $m = 5$, we have $\displaystyle\;\mathcal{E}_5 = \frac{25}{39366} \approx 6.35\times 10^{-4}\;$. This is already good enough for our purposes. Notice
$$\sum_{n=m}^{\infty}\frac{1}{n^3-n}
= \sum_{n=m}^{\infty}\frac{1}{(n-1)n(n+1)}
= \sum_{n=m}^{\infty}\left(\frac{1}{2(n-1)n}-\frac{1}{2n(n+1)}\right)
= \frac{1}{2(m-1)m}
$$
We find the original sum is approximately equal to 
$$-\frac34 \left( 1 + \frac{1}{2^3} + \frac{1}{3^3} + \frac{1}{4^3} + \frac{1}{2\cdot 4\cdot5} \right) = -\frac{10391}{11520} \approx -0.90199653$$
with an error smaller than $5\times 10^{-4}$.
Compare this with the exact value of the sum $\approx -0.90154268$, the difference
$\approx -4.5385 \times 10^{-4}$ is indeed smaller than $5 \times 10^{-4}$.
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