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

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
$$\boxed{\mathrm{Required\,Sum} = \sum_{n=2}^{\infty} \frac{-2}{(4n)^2  - 1} = \frac 1 2 \left( \frac \pi 2 - \frac {26}{15}\right)}$$
