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Find the limit of the following series:

$ 1 - \frac{1}{4} + \frac{1}{6} - \frac{1}{9} + \frac{1}{11} - \frac{1}{14} + \cdot \cdot \cdot $

If i go the integration way all is fine for a while but then things become pretty ugly. I'm trying to find out if there is some easier way to follow.

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    @Egbert:Very nice.2012-05-29

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Below are my personal notes on a powerful method for the summation of series using complex analytic methods. The result is quite general and the final calculation is simple. If you keep this up your sleeve, later you'll come across this problem again and in Ramanujan-like fashion immediately state "The value of this sum is simply $ \sum_{\zeta= \pm 1/5} \operatorname{Res}_{z=\zeta}\left( \frac{\pi \cot(\pi z)}{25z^2-1} \right) = \frac{\pi}{5} \cot \left( \frac{\pi}{5} \right)."$

P.S. Sorry it isn't LaTeXed up, I don't have the stamina to type all this up at the moment.

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    @Ragib Zaman: nice. It would be great if you found a time window to LaTeX it up.2012-05-29
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$1 - \lim\limits_{n\to\infty}\sum\limits_{i=1}^{n}{\frac{2}{(25i^2-1)}}$

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    I fixed that typo.2012-05-29
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Let $S = 1 - x^{3} + x^{5} -x^{8} + x^{10} - x^{13} + \cdots$. Then what you want is $\int_{0}^{1} S \ dx$. But we have \begin{align*} S &= 1 - x^{3} + x^{5} -x^{8} + x^{10} - x^{13} + \cdots \\\ &= -(x^{3}+x^{8} + x^{13} + \cdots) + (1+x^{5} + x^{10} + \cdots) \\\ &= -\frac{x^{3}}{1-x^{5}} + \frac{1}{1-x^{5}} \end{align*}

Now you have to evaluate: $\displaystyle \int_{0}^{1}\frac{1-x^{3}}{1-x^{5}} \ dx$

And wolfram gives the answer as: enter image description here

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    no. Please let it here. Don't delete it.2012-05-29