FOUR BEAUTIFUL INFINITE SERIES RELATED TO THE COSECANT EXPANSION

     Today´s post I want to proof the following  beautiful results


\huge\sum_{n=-\infty}^{\infty}\frac{(-1)^n}{16n^2-1}=-\frac{\pi \sqrt{2}}{4}

\huge\sum_{n=1}^{\infty}\frac{(-1)^n}{16n^2-1}=\frac{1}{2}-\frac{\pi \sqrt{2}}{8}

\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}=\frac{\pi^2 \sqrt{2}}{32}+\frac{\pi\sqrt{2}}{8 }

\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}= -\frac{1}{2}+\frac{\pi^2 \sqrt{2}}{64}+\frac{\pi\sqrt{2}}{16 }

     Let´s start from the first one. Note that we can break down this series in three pieces


\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{(4n)^2-1}=\sum_{n=- \infty}^{-1}\frac{(-1)^n}{(4n)^2-1}+\frac{(-1)^0}{(4\cdot0)^2-1}+\sum_{n=1}^{\infty}\frac{(-1)^n}{(4n)^2-1}


letting n=-m in the first sum of the R.H.S. we obtain

\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{(4n)^2-1}=\sum_{m= \infty}^{1}\frac{(-1)^{-m}}{(4(-m))^2-1}+\frac{(-1)^0}{(4\cdot0)^2-1}+\sum_{n=1}^{\infty}\frac{(-1)^n}{(4n)^2-1}

\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{(4n)^2-1}=-1+2\sum_{n=1}^{\infty}\frac{(-1)^n}{(4n)^2-1}(1)

\sum_{n=1}^{\infty}\frac{(-1)^n}{(4n)^2-1}=\frac{1}{16}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2-\frac{1}{16}}(2)

Recall the series expansion for \csc\left(\pi x \right) proved here

\sum_{n=1}^{\infty}\frac{(-1)^n}{x^2-n^2}=\frac{\pi\csc\left(\pi x \right)}{2x}-\frac{1}{2x^2}

\Rightarrow\,\, \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2-x^2}=\frac{1}{2x^2}-\frac{\pi\csc\left(\pi x \right)}{2x}(3)

comparing (2) and (3) and equating  x=\frac{1}{4}  we obtain

\sum_{n=1}^{\infty}\frac{(-1)^n}{(4n)^2-1}=\frac{1}{16}\left(8-2\pi\csc\left(\frac{\pi}{4}\right) \right)


\boxed{\sum_{n=1}^{\infty}\frac{(-1)^n}{(4n)^2-1}=\frac{1}{2}-\frac{\pi \sqrt{2}}{8}}(4)

Pluging (4) in (1) we get

\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{(4n)^2-1}=-1+2\left(\frac{1}{2}-\frac{\pi \sqrt{2}}{8}\right)


\boxed{\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{(4n)^2-1}=-\frac{\pi \sqrt{2}}{4}}(5)


And we are done with the first two!

For the last two We can differentiate (1) with respect to x:


2x\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(n^2-x^2\right)^2}=\frac{d}{dx}\left(\frac{1}{2x^2}-\frac{\pi\csc\left(\pi x \right)}{2x}\right)

2x\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(n^2-x^2\right)^2}=-\frac{1}{x^3}-\frac{d}{dx}\left(\frac{\pi\csc\left(\pi x \right)}{2x}\right)

2x\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(n^2-x^2\right)^2}=-\frac{1}{x^3}-\frac{\pi}{2}\left(\frac{\pi x\csc^{\prime}\left(\pi x \right)-\csc\left(\pi x \right)}{x^2}\right)

2x\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(n^2-x^2\right)^2}=-\frac{1}{x^3}-\frac{\pi}{2}\left(\frac{-\pi x\cot\left(\pi x \right)\csc\left(\pi x \right)-\csc\left(\pi x \right)}{x^2}\right)

2x\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(n^2-x^2\right)^2}=-\frac{1}{x^3}+\frac{\pi^2}{2}\left(\frac{ \cot\left(\pi x \right)\csc\left(\pi x \right)}{x}\right)+\frac{\pi\csc\left(\pi x \right)}{2 x^2}


\boxed{\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(n^2-x^2\right)^2}=-\frac{1}{2x^4}+\frac{\pi^2}{4}\left(\frac{ \cot\left(\pi x \right)\csc\left(\pi x \right)}{x^2}\right)+\frac{\pi\csc\left(\pi x \right)}{4 x^3}}(6)

Then we get

\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}=\sum_{n=- \infty}^{-1}\frac{(-1)^n}{\left(16n^2-1\right)^2}+1+\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}

Letting n=-m in the first sum of the R.H.S. we get

\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}=\sum_{m= \infty}^{1}\frac{(-1)^{-m}}{\left(16(-m)^2-1\right)^2}+1+\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}

\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}=1+2\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}(7)

Lets focus now in

\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}=\frac{1}{256}\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(n^2-\frac{1}{16}\right)^2}

From (6) we get

\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(n^2-x^2\right)^2}=-\frac{1}{2x^4}+\frac{\pi^2}{4}\left(\frac{ \cot\left(\pi x \right)\csc\left(\pi x \right)}{x^2}\right)+\frac{\pi\csc\left(\pi x \right)}{4 x^3}

\sum_{n=1}^{\infty}\frac{(-1)^n}{\left((4n)^2-1\right)^2}=\frac{1}{256}\left( -\frac{256}{2}+\frac{\pi^2}{4}\left(16 \cot\left(\frac{\pi}{4} \right)\csc\left(\frac{\pi}{4}\right)\right)+\frac{64\pi\csc\left(\frac{\pi}{4} \right)}{4 }\right)

=\left( -\frac{1}{2}+\frac{\pi^2}{64}\left(\cot\left(\frac{\pi}{4} \right)\csc\left(\frac{\pi}{4}\right)\right)+\frac{\pi\csc\left(\frac{\pi}{4} \right)}{16 }\right)

And finally


\boxed{\sum_{n=1}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}= -\frac{1}{2}+\frac{\pi^2 \sqrt{2}}{64}+\frac{\pi\sqrt{2}}{16 }}(8)

Plugging (8) in (7)


\boxed{\sum_{n=- \infty}^{\infty}\frac{(-1)^n}{\left(16n^2-1\right)^2}=\frac{\pi^2 \sqrt{2}}{32}+\frac{\pi\sqrt{2}}{8 }}(9)





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