How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue)
$$
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$ $$
1 + \sum\limits_{n = 0}^\infty  {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)}  = 1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$
I ask
$$
f\left( z \right) =  - \frac{2}{{\left( {4z + 9} \right)\left( {4z + 7}\right)}}
$$
is to :
$$\sum\limits_{n =  - \infty }^\infty  {\frac{2}{{\left( {4n + 9} \right)\left( {4n + 7}\right)}}}  = \left( {\mathop {\lim }\limits_{z \to  - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] + \mathop {\lim }\limits_{z \to  - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}}\right] } \right)$$
I found:
\begin{array}{l}
 \mathop {\lim }\limits_{z \to  - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to  - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{9}{4}} \right)\left( {4z + 7} \right)}}} \right] \\ 
 \quad \quad \quad \quad \quad \quad \quad \quad \quad  = \frac{1}{4}\mathop {\lim }\limits_{z \to  - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{4z + 7}}} \right] = \frac{1}{4}\left[ {\frac{{ - \pi }}{{ - 2}}} \right] = \frac{\pi }{8} \\ 
 \mathop {\lim }\limits_{z \to  - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to  - \frac{9}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{7}{4}} \right)\left( {4z + 9} \right)}}} \right] \\ 
 \quad \quad \quad \quad \quad \quad \quad \quad \quad  = \frac{1}{4}\mathop {\lim }\limits_{z \to  - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)}}} \right] = \frac{\pi }{8} \\ 
 \end{array}
\begin{array}{l}
 \sum\limits_{n =  - \infty }^\infty  {\frac{2}{{\left( {4n + 9} \right)4n + 7}}}  =  - \frac{\pi }{4} =  - \left( {\frac{\pi }{8} + \frac{\pi }{8}} \right) \\ 
  \Rightarrow s = 1 + \sum\limits_{n = 0}^\infty  {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)}  = 1 - \frac{\pi }{8} = \frac{{7 - \pi }}{8} \\ 
 \end{array}
I have a question for the result 
$$\sum\limits_{n = 0}^\infty  {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)}  =  - \frac{1}{5} \Rightarrow s = 1 + \sum\limits_{n = 0}^\infty  {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)}  = \frac{4}{5} \ne \frac{{7 - \pi }}{8}$$
thank you in advance
 A: Here is a method without complex analysis. I use the following two:
$$\int_0^1 x^{4n+8}\,dx=\frac{1}{4n+9}$$
$$\int_0^1 x^{4n+6}\,dx=\frac{1}{4n+7}$$
to get:
$$\sum_{n=0}^{\infty} \left(\frac{1}{4n+9}-\frac{1}{4n+7}\right)=\int_0^1 \sum_{n=0}^{\infty} \left(x^{4n+8}-x^{4n+6}\right)\,dx=\int_0^1 \frac{x^8-x^6}{1-x^4}\,dx$$
$$\Rightarrow \int_0^1 \frac{x^8-x^6}{1-x^4}\,dx=\int_0^1 \frac{-x^6}{1+x^2}\,dx=-\left(\int_0^1 \frac{1+x^6-1}{1+x^2}\,dx\right)$$
$$=-\int_0^1 \frac{1+x^6}{1+x^2}\,dx+\int_0^1 \frac{1}{1+x^2}\,dx$$
Write $1+x^6=(1+x^2)(1-x^2+x^4)$ to obtain:
$$-\int_0^1 (x^4-x^2+1)\,dx+\int_0^1 \frac{1}{1+x^2}\,dx$$
Both the integrals are easy to evaluate, hence the final answer is:
$$\boxed{\dfrac{\pi}{4}-\dfrac{13}{15}}$$
A: (Posted as an answer in case my earlier comment is removed)
It is relatively easy to prove (either through elementary means or via complex analysis) the well-known identity $$1 + 2 \sum_{n=1}^\infty \frac{z^2}{z^2 - (n\pi)^2} = z \cot z.$$  Then the given series (not the one written in summation notation, but the one that was actually written out) $$1 - \frac{1}{7} + \frac{1}{9} + \frac{1}{15} - \frac{1}{17} + \cdots$$ is simply the special case $z = \frac{\pi}{8}$.
A: $1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - \frac{1}{{23}} + \frac{1}{{25}} - .... = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 1}}}  - \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 7}}} $
\begin{array}{l}
 1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - \frac{1}{{23}} + \frac{1}{{25}} - .... = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 1}}}  - \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 7}}}  \\ 
 f\left( z \right) = \frac{1}{{8z + 1}} - \frac{1}{{8z + 7}} = \frac{6}{{\left( {8n + 1} \right)\left( {8n + 7} \right)}} \\ 
 \frac{1}{8}\mathop {\lim }\limits_{z \to  - \frac{1}{8}} \left( {z + \frac{1}{8}} \right)\frac{{6\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{1}{8}} \right)\left( {8z + 7} \right)}} = \frac{\pi }{8}\cot \left( { - \frac{\pi }{8}} \right) =  - \frac{{\pi \left( {1 + \sqrt 2 } \right)}}{8} \\ 
 \frac{1}{8}\mathop {\lim }\limits_{z \to  - \frac{1}{8}} \left( {z + \frac{7}{8}} \right)\frac{{6\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{7}{8}} \right)\left( {8z + 1} \right)}} =  - \frac{\pi }{8}\cot \left( { - \frac{{7\pi }}{8}} \right) =  - \frac{{\pi \left( {1 + \sqrt 2 } \right)}}{8} \\ 
 \sum\limits_{ - \infty }^{ + \infty } {\frac{6}{{\left( {8n + 1} \right)\left( {8n + 7} \right)}}}  =  - \left( {Residu\left( {\pi f\left( z \right)\cot \left( {\pi z} \right); - \frac{1}{8}} \right) + Re sidu\left( {\pi f\left( z \right)\cot \left( {\pi z} \right); - \frac{7}{8}} \right)} \right) \\ 
  \Rightarrow \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 1}}}  - \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 7}}}  = \frac{{\pi \left( {1 + \sqrt 2 } \right)}}{8} \\ 
 \end{array}
A: Here is a way to evaluate your series with the method of residues.
$$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = \sum_{n=2}^{\infty} \left(\frac{1}{4n+1} - \frac{1}{4n-1}\right) =\sum_{n=2}^{\infty} \frac{-2}{(4n)^2  - 1} = f(n)$$
Consider a function
$$  f(z)= \frac{-2}{(4z)^2  - 1} $$
Now, 
$$\sum_{n=-\infty}^{\infty} f(n) = 2\sum_{n=2}^{\infty} \frac{-2}{(4n)^2  - 1} + f(1)+f(0)+f(-1)$$
Using residue theorem we calculate the sum of residue as , 
$$\sum_{n=-\infty}^{\infty} f(n) = \frac \pi 2$$
and 
$$f(-1) + f(0) + f(1) = \frac{26}{15}$$
Putting it together you get
$$\boxed{\mathrm{Required\,Sum} = \sum_{n=2}^{\infty} \frac{-2}{(4n)^2  - 1} = \frac 1 2 \left( \frac \pi 2 - \frac {26}{15}\right)}$$
