How to find the sum of the following series ? Kindly guide me about the general term, then I can give a shot at summing it up.
$$1 - \frac{1}{4} + \frac{1}{6} -\frac{1}{9} +\frac{1}{11} - \frac{1}{14} + \cdots$$
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Your sum is $$S = \dfrac11 - \dfrac14 + \dfrac16 - \dfrac19 + \dfrac1{11} - \dfrac1{14} \pm \cdots = \sum_{k=0}^{\infty} \left(\dfrac1{5k+1} - \dfrac1{5k+4}\right)$$
Consider $f(t) = 1 - t^3 + t^5 - t^8 \pm \cdots $ for $\vert t \vert < 1$.
This is a geometric series and can be summed for $\vert t \vert < 1$.
Summing it up we get that $f(t) = \dfrac{1-t^3}{1-t^5}$ for $\vert t \vert <1$.
Now integrate $f(t)$ term by term from $0$ to $1$ and integrate $\dfrac{1-t^3}{1-t^5}$ from $0$ to $1$ to get the answer i.e. we have $$\sum_{k=0}^{\infty} \left(\dfrac1{5k+1} - \dfrac1{5k+4}\right) = \int_0^1 \dfrac{1-t^3}{1-t^5} dt$$
A similar sum using the same idea is worked out here.
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Telescopic series, isn't it? – Bazinga Jun 26 '12 at 05:10
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It is not a telescopic series, there is no cancellation occuring when you add two terms. – Jun 26 '12 at 05:15
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Sorry.Yes, its not. – Bazinga Jun 26 '12 at 05:17
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It appears that you're looking at $$\sum_{k=0}^\infty\left(\frac{1}{5k+1}-\frac{1}{5k+4}\right).$$
Cameron Buie
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Using the principal value for the doubly infinite harmonic series yields $$ \begin{align} \sum_{k=0}^{\infty} \left(\dfrac1{5k+1} - \dfrac1{5k+4}\right) &=\frac15\sum_{k=-\infty}^\infty\frac{1}{k+1/5}\\ &=\frac{\pi}{5}\cot\left(\frac{\pi}{5}\right)\\ &=\frac{\pi}{5}\sqrt{\frac{5+2\sqrt{5}}{5}} \end{align} $$
robjohn
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