What is the process for solving for the truncation error for a series?

Question

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

Answer 1

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 $$\sum_{n=101}^\infty\frac1{n^2} >\sum_{n=101}^\infty\frac1{n^2+n}=\frac1{101}. $$

Answer 2

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 $0.00995016666333415$..

Answer 3

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 $0.0099501666\ldots$.