TITLE: Infinite series with $2$ positive and $2$ negative terms
QUESTION [5 upvotes]: Finding value  of $$1+\frac{1}{5}-\frac{1}{7}-\frac{1}{11}+\frac{1}{13}+\frac{1}{17}-\frac{1}{19}-\frac{1}{23}+\cdots$$

Try: I have solved in using Integration
But i am trying to solver it without using integration  
Witting above series as $$\sum^{\infty}_{n=0}(-1)^n\bigg[\frac{1}{6n+1}+\frac{1}{6n+5}\bigg]$$
Now i am trying to solve it using euler reflection formula
Could not find any clue
could some help me to solve it

REPLY [5 votes]: $$1+\frac15-\frac17-\frac1{11}+\frac1{13}+\frac1{17}-\frac1{19}-\frac1{23}+\cdots=\frac{\pi}3$$

Utilizing the observation done by Mohammad Zuhair Khan we can explicitly write down the series as
$$1+\sum_{n=1}^\infty(-1)^n\left[\frac1{6n+1}-\frac1{6n-1}\right]=1-2\sum_{n=1}^\infty \frac{(-1)^n}{36n^2-1}$$
The latter form can be rewritten such that we can apply a sum identity of the cosecant function. To be precise we will use the formula

$$\pi\csc(\pi z)=z\sum_{n=-\infty}^\infty \frac{(-1)^n }{z^2-n^2}=\frac1z-2z\sum_{n=1}^\infty \frac{(-1)^n}{n^2-z^2}$$

Thus, lets rewrite the given series in the following way
$$\begin{align}
1-2\sum_{n=1}^\infty\frac{(-1)^n}{36n^2-1}&=1+\frac16\left[-2\frac16\sum_{n=1}^\infty \frac{(-1)^n}{n^2-\left(\frac16\right)^2}\right]\\
&=1+\frac16\left[\pi\csc\left(\frac\pi6\right)-6\right]\\
&=\frac\pi6\csc\left(\frac\pi6\right)
\end{align}$$

$$\therefore~1-2\sum_{n=1}^\infty \frac{(-1)^n}{36n^2-1}=\frac\pi3$$