What is the appropriate method to find the value of $1$ - $1\over 7$ + $1\over 13$ - ... upto infinite terms? What is the appropriate method to find the value of $1$ - $1\over 7$ + $1\over 13$ - ... upto infinite terms?
(The denominators increase by 6 in consecutive terms)
I approximated it by integrating $\frac{1}{1+x^6}$ putting x=1...is there a better and nicer method ? :)But what should I take as limits of the integration?
 A: For first, we have:
$$\begin{eqnarray*}\sum_{n\geq 0}\frac{(-1)^n}{6n+1}&=&\sum_{n\geq 0}(-1)^n\int_{0}^{1}x^{6n}\,dx=\int_{0}^{1}\frac{dx}{1+x^6}\tag{1}\end{eqnarray*}$$
Now we may compute the last integral through partial fraction decomposition. 
If $\xi_i$, $1\leq i\leq 6$, is a root of $1+x^6$, we have:
$$\text{Res}\left(\frac{1}{x^6+1},x=\xi_i\right)=\frac{1}{6\xi_i^5}=-\frac{\xi_i}{6}\tag{2}$$
hence:
$$ \int_{0}^{1}\frac{dx}{1+x^6}=-\frac{1}{6}\sum_{i=1}^{6}\int_{0}^{1}\frac{\xi_i}{x-\xi_i}\,dx=-\frac{1}{6}\sum_{i=1}^{6}\xi_i \log\left(1-\frac{1}{\xi_i}\right)\tag{3} $$
and:
$$\begin{eqnarray*} \int_{0}^{1}\frac{dx}{1+x^6}&=&-\frac{1}{6}\sum_{j=0}^{5}e^{\frac{\pi i}{6}(2j+1)}\log\left(1-e^{-\frac{\pi i}{6}(2j+1)}\right)\\&=&-\frac{1}{6}\sum_{j=0}^{5}e^{\frac{\pi i}{6}(2j+1)}\left(-\frac{\pi i}{12}(2j+1)+\log\left(2i\sin\frac{\pi(2j+1)}{12}\right)\right)\\&=&\frac{\pi}{6}-\frac{1}{6}\sum_{j=0}^{5}e^{\frac{\pi i}{6}(2j+1)}\left(\frac{\pi}{2}i+\log\sin\frac{\pi(2j+1)}{12}\right)\\&=&\color{red}{\frac{\pi+\sqrt{3}\log(2+\sqrt{3})}{6}}.\tag{4}\end{eqnarray*}$$

Another possible approach is the following: we have
$$ \sum_{n\geq 0}\frac{(-1)^n}{6n+1}=\sum_{n\geq 0}\left(\frac{1}{12n+1}-\frac{1}{12n+7}\right)=\frac{\psi\left(\frac{7}{12}\right)-\psi\left(\frac{1}{12}\right)}{12}\tag{5}$$
then the result follows from combining the reflection formula:
$$ \psi(z)-\psi(1-z)=-\pi\cot(\pi z)\tag{6}$$
with the duplication formula:
$$ \psi(z)+\psi\left(z+\frac{1}{2}\right)=-2\log 2+2\,\psi(2z)\tag{7}$$
and the triplication formula:
$$ 3\,\psi(3z)=(3\log 3)z+\psi(z)+\psi\left(z+\frac{1}{3}\right)+\psi\left(z+\frac{2}{3}\right)\tag{8}$$
for the digamma function.
A: I really like the approach that you would have liked to follow. 
Here I give a yet another approach which might not be the most straightforward way. Taking a look here helps us a bit:
\begin{align}
\sum_{n=0}^{\infty}\frac{(-1)^{n}}{1+6n}&=\frac16\sum_{n=0}^{\infty}\frac{(\color{blue}{-1})^{n}}{(n+\color{green}{\frac16})^\color{red}1}\\
&=\frac16\Phi(\color{blue}{-1},\color{red}1,\color{green}{\frac16})\\
&=\frac16\frac{1}{\Gamma(\color{red}1)}\int_0^{\infty}\frac{t^{\color{red}1-1}e^{-\color{green}{\frac16}t}}{1-(\color{blue}{-1})e^{-t}}dt\\
&=\frac16\int_0^{\infty}\frac{e^{-t/6}}{1+e^{-t}}dt\\
&=\int_0^{\infty}\frac{e^{-t}}{1+e^{-6t}}dt\\
&=\int_1^{\infty}\frac{t^4}{1+t^6}dt
\end{align}
where the last integral (which is manageable) results from $t\to \log t$.

Hint on solving the integral:
\begin{align}
\frac{t^4}{1+t^6}&=\frac{t^4}{(t^2+1)(t^4-t^2+1)}\\
&=\frac13\frac{1}{t^2+1}+\frac13\frac{2t^2-1}{t^4-t^2+1}\\
&=\frac13\frac{1}{t^2+1}+\frac13\frac{2t^2-1}{(t^2+1)^2-3t^2}\\
&=\frac13\frac{1}{t^2+1}+\frac13\frac{2t^2-1}{(t^2+1-\sqrt3t)(t^2+1+\sqrt3t)}\\
\end{align}
A: Define f(x) = x - $(x^7)\over 7$ +$ (x^{13})\over 13$ - ...
Note that f($0$)= $0$ and f($1$) is what you want to find.
You've already gathered that the derivative f'(x) = $1\over (1+x^6)$
$$  f(x)=\int_{?}^{x}\frac{1}{1+x'^{6}}dx'  $$
Your lower limit needs to be $0$ so that f($0$) will come out to $0$. Your upper limit needs to be $1$ because f($0$) is what you're trying to find.
$$    \sum_{n=0}^{\infty}\frac{1}{1+6n}\left(-1\right)^{n}=\int_{0}^{1}\frac{1}{1+x'^{6}}dx'
    $$
