How to estimate $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}$ with error less than $0.01$? How to estimate $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}$ with error less than $0.01$?
In order to solve the question, I think we need to write out the terms. 
So $\sum_{n=1}^{\infty}\frac{(-1)^n}{n^3}=-1+\frac18-\frac{1}{27}+\frac{1}{64}-...+\frac{(-1)^n}{n^3}$. However I don't see the pattern, so how are we going to estimate it?
 A: Short answer: the fifth term is the first term with absolute value less than $0.01$, so the sum can be estimated within $0.01$ of the actual value with only the first $4$ terms!

We know that this series converges, because it is an alternating series whose terms are strictly decreasing and go to zero. 
Let $S$ be the value of the sum. Let $S_k$ be the $k^{\text{th}}$ partial sum, or 
$$S_k=\displaystyle\sum_{n=1}^{k} \frac{(-1)^n}{n^3}$$
The fact that the series is alternating and has strictly decreasing terms means that 
$$
S_k > S \qquad\qquad \text{if } k \equiv 0 \pmod{2}
$$
and
$$
S_k < S \qquad\qquad \text{if } k \equiv 1 \pmod{2}
$$

This implies, for all $k$,
$$|S-S_k| < \frac{1}{(k+1)^3}$$
The first $k$ for which 
$$\frac{1}{(k+1)^3} < \frac{1}{100}$$
is $k=4$. Therefore you can estimate with
$$S \approx \sum_{n=1}^4 \frac{(-1)^n}{n^3}$$
A: The error in an alternating series with decreasing term is no bigger than the first omitted term.
A: Hint. One may observe that, as any alternating series with decreasing general term, one has
$$
\left|\sum_{n=N}^{\infty}\frac{(-1)^n}{n^3}\right|\leq \left|\frac{(-1)^N}{N^3}\right|,
$$ then we would like to have
$$
\frac1{N^3}< \frac{0.01}2
$$ that is
$$
N\geq 6.
$$ Then one may evaluate
$$
\sum_{n=1}^{6}\frac{(-1)^n}{n^3}
$$ with error less than $\dfrac{0.01}2$.
A: Each term generated by $(-1)^n/n^3$ adds to (or subtracts from) the sum. In order to find the sum to within 0.01, or $1/100$, we need to find when the absolute value of the terms are less than $1/100$.
\begin{align}
\frac{1}{n^3}&<\frac{1}{100}\\
100&<n^3\\
n&>\sqrt[3]{100}\approx4.64
\end{align}
Thus, the sum will be within 0.01 after 4 terms. Search on Alternating Series Remainder for more information.
