How to find the sum of this infinite series. 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.
$$1 - \frac{1}{4} + \frac{1}{6} -\frac{1}{9} +\frac{1}{11} - \frac{1}{14} + \cdots$$
 A: 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.
A: It appears that you're looking at $$\sum_{k=0}^\infty\left(\frac{1}{5k+1}-\frac{1}{5k+4}\right).$$
A: Using the principal value for the doubly infinite harmonic series yields
$$
\begin{align}
\sum_{k=0}^{\infty} \left(\dfrac1{5k+1} - \dfrac1{5k+4}\right)
&=\frac15\sum_{k=-\infty}^\infty\frac{1}{k+1/5}\\
&=\frac{\pi}{5}\cot\left(\frac{\pi}{5}\right)\\
&=\frac{\pi}{5}\sqrt{\frac{5+2\sqrt{5}}{5}}
\end{align}
$$
