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I have the following series:

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

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

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Writing the $n$th 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 $$\frac{1}{n - 1} = \int_0^1 x^{n - 2} \, dx \qquad \text{and} \qquad \frac{1}{n + 2} = \int_0^1 y^{n + 1} \, dy,$$ the sum can be rewritten as \begin{align*} \sum_{n = 2}^\infty \frac{(-1)^n}{n^2 + n - 2} &= \sum_{n = 2}^\infty \frac{(-1)^n}{(n - 1)(n + 2)}\\ &= \int_0^1 \int_0^1 \sum_{n = 2}^\infty (-1)^n x^{n - 2} y^{n + 1} \, dx dy \tag1\\ &= \int_0^1 \int_0^1 \frac{y}{x^2} \sum_{n = 2}^\infty (-xy)^n \, dx dy\\ &= \int_0^1 \int_0^1 \frac{y}{x^2} \cdot \frac{x^2 y^2}{1 + xy} \, dx dy \tag2\\ &= \int_0^1 \int_0^1 \frac{y^3}{1 + xy} \, dx dy\\ &= \int_0^1 y^2 \Big{[} \ln (1 + xy) \Big{]}_0^1 \, dy\\ &= \int_0^1 y^2 \ln (1 + y) \, dy\\ &= \frac{1}{3} \ln (2) - \frac{1}{3} \int_0^1 \frac{y^3}{1 + y} \, dy \tag3\\ &= \frac{1}{3} \ln (2) - \frac{1}{3} \int_0^1 \left (y^2 - y + 1 - \frac{1}{1 + y} \right ) \, dy \tag4\\ &= \frac{1}{3} \ln (2) - \frac{1}{3} \left [\frac{y^3}{3} - \frac{y^2}{2} + y - \ln (1 + y) \right ]_0^1\\ &= \frac{2}{3} \ln (2) - \frac{5}{18}. \end{align*}

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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  • $\begingroup$ Hello, omegadot. I liked the way you solved this sum problem. I am thinking whether this method of summing can be applied in other situations also. $\endgroup$ – RAHUl JHa Jan 30 '18 at 7:35
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    $\begingroup$ Whenever I see linear factors in the denominator of a convergent sum and a numerator of either $1$ or $(-1)^n$ I immediately think of trying to write the $n$th term of the sum in terms of an integral. After interchanging the integral sign with the summation, summing the resultant series which is geometric can be readily done. One hopefully can then perform the remaining integration. I have recently done this for a number of examples which you can follow here. $\endgroup$ – omegadot Feb 5 '18 at 23:30
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    $\begingroup$ It is just another tool to add to your toolbox. The only potential problem the method may throw up is integrals which are either tedious or difficult to find. $\endgroup$ – omegadot Feb 6 '18 at 3:53
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    $\begingroup$ Yes. But remember the sum this can be done on must first converge (try the technique on $\sum_{n =1}^\infty 1/n$ as see what happens) and the number of linear factors appearing in the denominator should not be too many since as their number grows the number of integrations that need to be performed also grows and can become very tedious. $\endgroup$ – omegadot Feb 6 '18 at 4:04
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    $\begingroup$ And as an example of a sum that leads to a more difficult integral to find (unless you are familiar with special functions that is or other more advanced techniques for finding improper integrals) is $\sum_{n = 1}^\infty 1/n^2$. See what you end up with if you let $1/n = \int_0^1 x^{n - 1} \, dx$ and $1/n = \int_0^1 y^{n - 1} \, dy$. $\endgroup$ – omegadot Feb 6 '18 at 4:06
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You need to fully expand the partial fraction.

\begin{align} (-1)^n \over (n+2)(n-1) &= {((n+2) - (n-1)) \times (-1)^n \over 3\times (n+2)(n-1)} \\ &= {(n+2) \times (-1)^n \over 3\times (n+2)(n-1)} - {(n-1) \times (-1)^n \over 3\times (n+2)(n-1)} \\ &= {(-1)^n \over 3\times (n-1)} - {(-1)^n \over 3\times (n+2)} \end{align}

Now this can be handled using Alternating Harmonic Series.

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

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$$\sum_{k=2}^{+\infty}\frac{(-1)^n}{n^2+n-2}=\frac{1}{3}\sum_{k=2}^{+\infty}(-1)^n\left(\frac{1}{n-1}-\frac{1}{n+2}\right)=$$ $$=\frac{1}{3}\left(1-\frac{1}{4}-\frac{1}{2}+\frac{1}{5}+\frac{1}{3}-\frac{1}{6}-\frac{1}{4}+\frac{1}{7}+...\right)=$$ $$=-\frac{1}{3}\left(1-\frac{1}{2}+\frac{1}{3}\right)+\frac{2}{3}\sum_{k=1}^{+\infty}\frac{(-1)^{k+1}}{k}=\frac{2}{3}\ln2-\frac{5}{18}.$$

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