TITLE: Evaluate $\sum_{n=1}^{\infty}{1\over n^5}$ up to the second decimal place
QUESTION [5 upvotes]: 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?

REPLY [3 votes]: Alternating Series
$$
\begin{align}
&1+\frac1{2^5}+\frac1{3^5}+\frac1{4^5}+\frac1{5^5}+\frac1{6^5}+\dots\\
&\phantom{1}-\frac{2}{2^5}\phantom{+\frac1{3^5}1}-\frac2{4^4}\phantom{+\frac1{5^5}1}-\frac{2}{6^5}\tag*{$\left(-\frac2{2^5}\text{times the line above}\right)$}\\
=&1-\frac1{2^5}+\frac1{3^5}-\frac1{4^5}+\frac1{5^5}-\frac1{6^5}+\dots\\
\end{align}
$$
Thus, the alternating sum is $\frac{15}{16}$ of the non-alternating sum. By the Alternating Series Test, the error at any point is less than the first unused term. Thus to get two digits of precision, we only need to go up to, but not including, $\frac1{3^5}=\frac1{243}$. That is,
$$
\begin{align}
\sum_{k=1}^\infty\frac1{k^5}
&\approx\frac{16}{15}\left(1-\frac1{2^5}\right)\\
&=1.0333333
\end{align}
$$
to within $\frac{16}{15}\frac1{243}=0.0043896$. In fact, the Alternating Series Test says that
$$
1.0333333\le\sum_{k=1}^\infty\frac1{k^5}\le1.0377229
$$
where the upper bound is
$$
\frac{16}{15}\left(1-\frac1{2^5}+\frac1{3^5}\right)
$$
The next lower bound is
$$
\frac{16}{15}\left(1-\frac1{2^5}+\frac1{3^5}-\frac1{4^5}\right)=1.0366812
$$
So now we can say that
$$
\bbox[5px,border:2px solid #C0A000]{1.0366812\le\sum_{k=1}^\infty\frac1{k^5}\le1.0377229}
$$
Thus, to two decimal places, the sum would be $1.04$.

Euler-Maclaurin Sum Formula
Using the Euler-Maclaurin Sum Formula:
$$
\begin{align}
&\sum_{k=1}^{100}\frac1{k^5}-\left[-\frac1{4n^4}+\frac1{2n^5}-\frac5{12n^6}+\frac7{24n^8}-\frac1{2n^{10}}+\frac{11}{8n^{12}}-\frac{65}{12n^{14}}\right]_{n=100}\\[9pt]
&=1.036927755143369926331365486457
\end{align}
$$
The next term is $\frac{691}{24n^{16}}$, so the error is on the order of $3\times10^{-31}$.

Estimating the Error with Bernoulli's Inequality
Using the Alternating Series Test is pretty easy, but we can also use Bernoulli's Inequality to estimate the error. For $p\ge1$,
$$
\begin{align}
\frac1{n^p}-\frac1{(n+1)^p}
&=\frac1{n^p}\left[1-\frac1{\left(1+\frac1n\right)^p}\right]\\
&\ge\frac1{n^p}\left[1-\frac1{1+\frac pn}\right]\\
&=\frac1{n^p}\frac{\frac pn}{1+\frac pn}\\
&\ge\frac1{n^p}\frac{\frac pn}{\left(1+\frac1n\right)^p}\\[4pt]
&=\frac p{n(n+1)^p}\\[8pt]
&\ge\frac p{(n+1)^{p+1}}
\end{align}
$$
The first two inequalities are due to Bernoulli's inequality, the last is simply because $n\lt n+1$. Using this inequality, we get that
$$
\begin{align}
\sum_{k=1}^\infty\frac1{k^{p+1}}-\sum_{k=1}^n\frac1{k^{p+1}}
&=\sum_{k=n+1}^\infty\frac1{k^{p+1}}\\
&=\sum_{k=n}^\infty\frac1{(k+1)^{p+1}}\\
&\le\frac1p\sum_{k=n}^\infty\left(\frac1{k^p}-\frac1{(k+1)^p}\right)\\
&=\frac1{p\,n^p}
\end{align}
$$
Thus, the error in approximating the infinite sum using $n$ terms is at most $\frac1{4n^4}$. Therefore, the error in
$$
\sum_{k=1}^3\frac1{k^5}=1.0353652
$$
is at most $\frac1{4\cdot3^4}=0.0030864$. That is,
$$
\bbox[5px,border:2px solid #C0A000]{1.0353652\le\sum_{k=1}^\infty\frac1{k^5}\le1.0384516}
$$
Thus, to two decimal places, the sum would be $1.04$.