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

(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 $z = \frac{\pi}{8}$.
