How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $

Question

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

Answer 1

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: $$\boxed{\dfrac{\pi}{4}-\dfrac{13}{15}}$$

Answer 2

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

Answer 3

$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}

Answer 4

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