besnuffelfo

2022-09-21

sum $\sum _{n=2}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{{n}^{2}+n-2}$
I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:
$\sum _{n=2}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{\left(n+2\right)\left(n-1\right)}$
But nothing seems to cancel (as usually happens) using the telescoping method. How can I solve the above series? Is there any other method to do the above problem?

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rae2721

Expert

You need to fully expand the partial fraction.
$\begin{array}{rl}\frac{\left(-1{\right)}^{n}}{\left(n+2\right)\left(n-1\right)}& =\frac{\left(\left(n+2\right)-\left(n-1\right)\right)×\left(-1{\right)}^{n}}{3×\left(n+2\right)\left(n-1\right)}\\ & =\frac{\left(n+2\right)×\left(-1{\right)}^{n}}{3×\left(n+2\right)\left(n-1\right)}-\frac{\left(n-1\right)×\left(-1{\right)}^{n}}{3×\left(n+2\right)\left(n-1\right)}\\ & =\frac{\left(-1{\right)}^{n}}{3×\left(n-1\right)}-\frac{\left(-1{\right)}^{n}}{3×\left(n+2\right)}\end{array}$
Now this can be handled using Alternating Harmonic Series.
The final result, as Mathematica calculated, is $\frac{-5+12\mathrm{log}2}{18}$

Still Have Questions?

Expert

Writing the nth term of your sum in a form that telescopes can be avoid altogether by converting the problem to a double integral as follows.
Noting that
$\frac{1}{n-1}={\int }_{0}^{1}{x}^{n-2}\phantom{\rule{thinmathspace}{0ex}}dx\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\frac{1}{n+2}={\int }_{0}^{1}{y}^{n+1}\phantom{\rule{thinmathspace}{0ex}}dy,$
the sum can be rewritten as
$\begin{array}{rl}\sum _{n=2}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{{n}^{2}+n-2}& =\sum _{n=2}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{\left(n-1\right)\left(n+2\right)}\\ \text{(1)}& & ={\int }_{0}^{1}{\int }_{0}^{1}\sum _{n=2}^{\mathrm{\infty }}\left(-1{\right)}^{n}{x}^{n-2}{y}^{n+1}\phantom{\rule{thinmathspace}{0ex}}dxdy& ={\int }_{0}^{1}{\int }_{0}^{1}\frac{y}{{x}^{2}}\sum _{n=2}^{\mathrm{\infty }}\left(-xy{\right)}^{n}\phantom{\rule{thinmathspace}{0ex}}dxdy\\ \text{(2)}& & ={\int }_{0}^{1}{\int }_{0}^{1}\frac{y}{{x}^{2}}\cdot \frac{{x}^{2}{y}^{2}}{1+xy}\phantom{\rule{thinmathspace}{0ex}}dxdy& ={\int }_{0}^{1}{\int }_{0}^{1}\frac{{y}^{3}}{1+xy}\phantom{\rule{thinmathspace}{0ex}}dxdy\\ & ={\int }_{0}^{1}{y}^{2}\left[\mathrm{ln}\left(1+xy\right){\right]}_{0}^{1}\phantom{\rule{thinmathspace}{0ex}}dy\\ & ={\int }_{0}^{1}{y}^{2}\mathrm{ln}\left(1+y\right)\phantom{\rule{thinmathspace}{0ex}}dy\\ \text{(3)}& & =\frac{1}{3}\mathrm{ln}\left(2\right)-\frac{1}{3}{\int }_{0}^{1}\frac{{y}^{3}}{1+y}\phantom{\rule{thinmathspace}{0ex}}dy\text{(4)}& & =\frac{1}{3}\mathrm{ln}\left(2\right)-\frac{1}{3}{\int }_{0}^{1}\left({y}^{2}-y+1-\frac{1}{1+y}\right)\phantom{\rule{thinmathspace}{0ex}}dy& =\frac{1}{3}\mathrm{ln}\left(2\right)-\frac{1}{3}{\left[\frac{{y}^{3}}{3}-\frac{{y}^{2}}{2}+y-\mathrm{ln}\left(1+y\right)\right]}_{0}^{1}\\ & =\frac{2}{3}\mathrm{ln}\left(2\right)-\frac{5}{18}.\end{array}$
Explanation
1. Interchanging the summation with the double integration.
2. Summing the series which is geometric.
3. Integration by parts.
4. Partial fraction decomposition.

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