What is the process for solving for the truncation error for a series? Give an appropriate upper bound for the following truncation error:
$$\left| \sum_{n=1}^{\infty} \frac{1}{n^2} - \sum_{n=1}^{100} \frac{1}{n^{2}} \right|$$
Please explain the process for solving for the upper bound truncation.
Thank you
 A: The error is bounded by
$$\begin{align}\sum_{n=101}^\infty\frac1{n^2} 
&< \sum_{n=101}^\infty\frac1{n^2-n}\\
&=  \sum_{n=101}^\infty\left(\frac1{n-1}-\frac1n\right)\\
&=\lim_{N\to\infty}\left(\sum_{n=101}^N\left(\frac1{n-1}-\frac1n\right)\right)\\
&=\lim_{N\to\infty}\left(\sum_{n=101}^N\frac1{n-1}-\sum_{n=101}^N\frac1{n}\right)\\
&=\lim_{N\to\infty}\left(\sum_{n=100}^{N-1}\frac1{n}-\sum_{n=101}^N\frac1{n}\right)\\
&=\lim_{N\to\infty}\left(\frac1{100}-\frac1N\right)\\&=\frac1{100}.
\end{align}$$
We "wasted" something only in the first step - but was it much?
It turns out it wasn't, for we also have
$$\sum_{n=101}^\infty\frac1{n^2} >\sum_{n=101}^\infty\frac1{n^2+n}=\frac1{101}. $$
A: We can provide upper and lower bounds by using integrals.  Note that we have the estimates 
$$\int_{101}^{\infty} \frac{1}{x^2}\,dx \le \sum_{n=101}^\infty\frac{1}{n^2}\le \frac{1}{(101)^2}+ \int_{101}^{\infty} \frac{1}{x^2}\,dx\tag 1$$
Carrying out the integral in $(1)$ yields
$$\frac{1}{101}\le \sum_{n=101}^\infty \frac{1}{n^2}\le \frac{1}{101}+\frac{1}{101^2}\tag 2$$
The lower bound in $(2)$ is $\frac1{101}\approx 0.0099009900990099$ while the upper bound in $(2)$ is $\frac{1}{101}+\frac{1}{101^2}\approx 0.00999901970395059$.
The actual numerical error is roughly $0.00995016666333415$..  
A: In the first lines of this proof of Stirling's inequality it is shown by creative telescoping that
$$ \color{red}{\frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}-\frac{1}{30m^5}}\leq \sum_{n\geq m}\frac{1}{n^2} \leq \color{red}{\frac{1}{m}+\frac{1}{2m^2}+\frac{1}{6m^3}}$$
and you just need to plug in $m=101$ to get that the approximation error is $0.0099501666\ldots$.
