<s>[INST] Evaluate $\sum_{n=1}^{\infty}{1\over n^5}$ up to the second decimal place I am trying to evaluate
$$\sum_{n=1}^{\infty}{1\over n^5}$$
up to the second decimal place. While the series is convergent, I have no idea how to construct such a bound, preferably using basic properties of series and sequences. Any hints?
 [/INST] Considering the telescoping sum
$$\sum_{n=2}^{\infty} \left[ \frac{1}{n^2(n-1)^2}-\frac{1}{n^2(n+1)^2} \right]=\frac{1}{4}$$
Also $$\frac{1}{n^2(n-1)^2}-\frac{1}{n^2(n+1)^2}=\frac{4}{n^{5}}+O\left( \frac{1}{n^7} \right)$$
\begin{align*}
  \sum_{n=1}^{\infty} \frac{1}{n^5} &=
  1+\frac{1}{16}+\sum_{n=2}^{\infty}
  \left[
    \frac{1}{n^5}-\frac{1}{4n^2(n-1)^2}+\frac{1}{4n^2(n+1)^2}
  \right] \\
  &= \frac{17}{16}-\sum_{n=2}^{\infty} \frac{2n^2-1}{n^5(n^2-1)^2} \\
  &\approx \frac{17}{16}-\sum_{n=2}^{\color{red}{3}} \frac{2n^2-1}{n^5(n^2-1)^2} \\
  &= 1.03710 \ldots \\
  &= \zeta(5)-0.000173 \ldots
\end{align*}
Error bound
\begin{align*}
  E(N) &=\sum_{n=N+1}^{\infty} \frac{2n^2-1}{n^5(n^2-1)^2} \\ &<
  \sum_{n=N+1}^{\infty} \frac{2}{n^3(n^2-1)^2} \\
  &< \sum_{n=N+1}^{\infty} \int_{n}^{n+1} \frac{2}{x^3(x^2-1)^2} \, dx \\
  &= \frac{2N^2-1}{N^2(N^2-1)}-\ln \frac{N^4}{(N^2-1)^2}
\end{align*}
$$
\begin{array}{|c|c|}
\hline
N & E(N)< \\
\hline
2 & 0.007969 \ldots \\
3 & 0.000545 \ldots \\
4 & 0.000089 \ldots \\
5 & 0.000022 \ldots \\
\hline
\end{array}
$$
That's why $N=\color{red}{3}$ is enough for $2$ decimal places.
</s>