besnuffelfo

Answered

2022-09-21

sum $\sum _{n=2}^{\mathrm{\infty}}\frac{(-1{)}^{n}}{{n}^{2}+n-2}$

I am not able to do the telescoping process in the above series. I converted it into the following partial fraction:

$$\sum _{n=2}^{\mathrm{\infty}}\frac{(-1{)}^{n}}{(n+2)(n-1)}$$

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?

Answer & Explanation

rae2721

Expert

2022-09-22Added 8 answers

You need to fully expand the partial fraction.

$$\begin{array}{rl}\frac{(-1{)}^{n}}{(n+2)(n-1)}& =\frac{((n+2)-(n-1))\times (-1{)}^{n}}{3\times (n+2)(n-1)}\\ & =\frac{(n+2)\times (-1{)}^{n}}{3\times (n+2)(n-1)}-\frac{(n-1)\times (-1{)}^{n}}{3\times (n+2)(n-1)}\\ & =\frac{(-1{)}^{n}}{3\times (n-1)}-\frac{(-1{)}^{n}}{3\times (n+2)}\end{array}$$

Now this can be handled using Alternating Harmonic Series.

The final result, as Mathematica calculated, is $\frac{-5+12\mathrm{log}2}{18}$

Conrad Beltran

Expert

2022-09-23Added 1 answers

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

$$\begin{array}{rl}\sum _{n=2}^{\mathrm{\infty}}\frac{(-1{)}^{n}}{{n}^{2}+n-2}& =\sum _{n=2}^{\mathrm{\infty}}\frac{(-1{)}^{n}}{(n-1)(n+2)}\\ \text{(1)}& & ={\int}_{0}^{1}{\int}_{0}^{1}\sum _{n=2}^{\mathrm{\infty}}(-1{)}^{n}{x}^{n-2}{y}^{n+1}\phantom{\rule{thinmathspace}{0ex}}dxdy& ={\int}_{0}^{1}{\int}_{0}^{1}\frac{y}{{x}^{2}}\sum _{n=2}^{\mathrm{\infty}}(-xy{)}^{n}\phantom{\rule{thinmathspace}{0ex}}dxdy\\ \text{(2)}& & ={\int}_{0}^{1}{\int}_{0}^{1}\frac{y}{{x}^{2}}\cdot \frac{{x}^{2}{y}^{2}}{1+xy}\phantom{\rule{thinmathspace}{0ex}}dxdy& ={\int}_{0}^{1}{\int}_{0}^{1}\frac{{y}^{3}}{1+xy}\phantom{\rule{thinmathspace}{0ex}}dxdy\\ & ={\int}_{0}^{1}{y}^{2}{\textstyle [}\mathrm{ln}(1+xy){{\textstyle ]}}_{0}^{1}\phantom{\rule{thinmathspace}{0ex}}dy\\ & ={\int}_{0}^{1}{y}^{2}\mathrm{ln}(1+y)\phantom{\rule{thinmathspace}{0ex}}dy\\ \text{(3)}& & =\frac{1}{3}\mathrm{ln}(2)-\frac{1}{3}{\int}_{0}^{1}\frac{{y}^{3}}{1+y}\phantom{\rule{thinmathspace}{0ex}}dy\text{(4)}& & =\frac{1}{3}\mathrm{ln}(2)-\frac{1}{3}{\int}_{0}^{1}({y}^{2}-y+1-\frac{1}{1+y})\phantom{\rule{thinmathspace}{0ex}}dy& =\frac{1}{3}\mathrm{ln}(2)-\frac{1}{3}{[\frac{{y}^{3}}{3}-\frac{{y}^{2}}{2}+y-\mathrm{ln}(1+y)]}_{0}^{1}\\ & =\frac{2}{3}\mathrm{ln}(2)-\frac{5}{18}.\end{array}$$

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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