# How to find the sum of this infinite series. [duplicate]

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.

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

## marked as duplicate by Cameron Buie, Tom-Tom, Joel Reyes Noche, Tim Raczkowski, Lord_FarinSep 18 '15 at 19:28

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.

• Telescopic series, isn't it? – Bazinga Jun 26 '12 at 5:10
• It is not a telescopic series, there is no cancellation occuring when you add two terms. – user17762 Jun 26 '12 at 5:15
• Sorry.Yes, its not. – Bazinga Jun 26 '12 at 5:17

It appears that you're looking at $$\sum_{k=0}^\infty\left(\frac{1}{5k+1}-\frac{1}{5k+4}\right).$$

• Yes, precisely. Thank you. – Bazinga Jun 26 '12 at 5:21

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}