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?
 A: We may use Beuker-like integrals, for instance. If we set $\zeta(5)=\sum_{n\geq 1}\frac{1}{n^5}$, we have:
$$ 24\,\zeta(5) = \int_{0}^{1}\frac{\log(x)^4}{1-x}\,dx,\qquad \int_{0}^{1}\frac{x^4\log(x)^4}{1-x}\,dx =  -\frac{257875}{10368}+24\,\zeta(5)\tag{1}$$
but the function $g(x)=\frac{x^4\log^4(x)}{1-x}$ is non-negative and bounded by $\frac{1}{60}$ on the interval $(0,1)$, hence:
$$ \zeta(5) \approx \frac{257875}{248832}= \color{red}{1.036}3417888\ldots \tag{2} $$
A: If you're allowed to use the fact that $\sum1/n^2=\pi^2/6$, you can calculate when the tail of $\sum1/n^2$ is less than $0.01$, say
$$\sum_{n\geq m}1/n^2<0.01$$
Then $S=\sum_{n\geq m} 1/n^5$ is also less than $0.01$ .
A: Let $N$ be such that $\sum \limits _{k \ge 2^N} ^\infty \frac 1 {n^5} \le \epsilon$. Notice that
$$\sum _{k \ge 2^N} ^\infty \frac 1 {n^5} \le \sum _{m \ge 0} \ \sum _{k = 2^{N+m}} ^{2^{N+m+1}-1} \frac 1 {n^5} \le \sum _{m \ge 0} \ \sum _{k = 2^{N+m}} ^{2^{N+m+1}-1} \frac 1 {(2^{N+m})^5} = \sum _{m \ge 0} \frac {2^{N+m}} {(2^{N+m})^5} = \sum _{m \ge 0} \frac 1 {(2^{N+m})^4} = \\
\frac 1 {2^{4N}} \sum _{m \ge 0} \frac 1 {2^{4m}} = \frac 1 {2^{4N}} \frac 1 {1 - \frac 1 {16}} = \frac 1 {15} \frac 1 {16^{N-1}} .$$
It follows that in order to have the inequality that we have begun with, it is sufficient to impose that $\frac 1 {15} \frac 1 {16^{N-1}} \le \epsilon$, which means $16^{N-1} \ge \frac 1 {15 \epsilon}$, whence $N-1 \ge \log_{16} \frac 1 {15 \epsilon} = -\log_{16} (15\epsilon) = -\frac 1 4 \log_2 (15 \epsilon)$, so $N = \left[ 1 -\frac 1 4 \log_2 (15 \epsilon) \right] + 1$, where $[x]$ denotes the integer part of $x$ (the "floor" function, as some call it).
Now just choose $\epsilon = 10^{-3}$ (because you want the first 2 decimal places to be exact) in order to get $N = \left[ 1 -\frac 1 4 \log_2 (15 \cdot 10^{-3}) \right] + 1 = 3$. This means that
$$\sum _{n=1} ^{2^3-1} \frac 1 {n^5} = 1.\color{red}{03}6849\dots$$
has the first 2 decimal places correct.
A: 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.
A: HINT:
Recall from The Integral Test
$$\int_N^{M+1} \frac{1}{x^5}\,dx\le \sum_{n=N}^M\frac{1}{n^5}\le \frac{1}{N^5}+\int_N^M \frac{1}{x^5}\,dx$$
A: For $n>1$ we have $$1/n^5+1/(n+1)^5+1/(n+2)^5+...<$$ $$<\frac {1}{n^3}\left(\frac {1}{(n-1)n}+\frac {1}{n(n+1)}+\frac {1}{(n+1)(n+2)}+...\right)=$$ $$=\frac {1}{n^3}\left((\frac {1}{n-1}-\frac {1}{n})+(\frac {1}{n}-\frac {1}{n+1})+(\frac {1}{n+1}-\frac {1}{n+2})+...\right)=$$ $$=\frac {1}{n^3(n-1)}$$ because the sum of the first $k$ terms of the last infinite series above is $\frac {1}{n-1}-\frac {1}{n-1+k}.$
This is not as sharp an upper bound as can be found by deeper methods. We have $1.036<\sum_{n=1}^{n=5}n^{-5}<1.037$ and the sum of the remaining terms is less than $1/(6^3\cdot 5)=1/1080.$ So the sum rounded to 2 decimal places is $1.04$
