COMPUTING INFINITE SUMS RELATED TO THE COTANGENT EXPANSION

           I got inspired by this post from @infseriesbot, and decided to proof only one of the relations appearing in that list because the others follow automatically by just applying the exact same technique. But, in order  to turn this post more fun and interesting, I decided to see if  We could expand that list and proof some variations. For instance, expanding the range of summation from  -\infty\,\,<k<\,\,+\infty, and also evaluate  some higher order powers variation. Doing that, I obtained the following beautiful results:

\sum_{k=1}^{\infty}\frac{1}{9k^2-1}=\frac{1}{2}-\frac{\pi}{6 \sqrt{3}}(1)

\sum_{k=-\infty}^{\infty}\frac{1}{9k^2-1}=-\frac{\pi}{3 \sqrt{3}}(2)

\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^2}=\frac{\pi^2}{16}-\frac{1}{2}(3)

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

\sum_{k=1}^{\infty}\frac{1}{\left(9k^2-1\right)^2}=-\frac{1}{2}+\frac{ \pi^2 }{48} +\frac{ \pi}{12\sqrt{3}}(5)

\sum_{k=-\infty}^{\infty}\frac{1}{\left(9k^2-1\right)^2}=\frac{ \pi^2 }{24} +\frac{ \pi}{6\sqrt{3}}(6)

\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^3}=\frac{1}{2}-\frac{3\pi^2}{64}(7)

\sum_{k=-\infty}^{\infty}\frac{1}{\left(4k^2-1\right)^3}=-\frac{3\pi^2}{32}(8)

The starting point for the proofs is the partial fractions expansion for the \cot(\pi x) already proved here, namely

\sum_{k=1}^{\infty}\frac{1}{k^2-x^2}=\frac{1}{2x^2}-\frac{\pi \cot\left(\pi x \right)}{2x}(9)

To proof (1) and (2), lets first observe the fact


\sum_{k=-\infty}^{\infty}\frac{1}{9k^2-1}=\sum_{k=-\infty}^{-1}\frac{1}{9k^2-1}+\frac{1}{9\cdot(0)-1}+\sum_{k=1}^{\infty}\frac{1}{9k^2-1}


let k=-m in the first sum on the R.H.S., then


\sum_{k=-\infty}^{\infty}\frac{1}{9k^2-1}=-1+\sum_{m=\infty}^{1}\frac{1}{9(-m)^2-1}+\sum_{k=1}^{\infty}\frac{1}{9k^2-1}

\sum_{k=-\infty}^{\infty}\frac{1}{9k^2-1}=-1+2\sum_{k=1}^{\infty}\frac{1}{9k^2-1}(10)

Then

\sum_{k=1}^{\infty}\frac{1}{9k^2-1}=\frac{1}{9}\sum_{k=1}^{\infty}\frac{1}{k^2-\frac{1}{9}}

Letting x=\frac{1}{3} in (9) we get that

\sum_{k=1}^{\infty}\frac{1}{9k^2-1}=\frac{1}{9}\left(\frac{9}{2}-\frac{3\pi }{2}\cot\left(\frac{\pi}{3} \right) \right)

=\frac{1}{2}-\frac{\pi }{6}\frac{\sqrt{3}}{3}

\boxed{\sum_{k=1}^{\infty}\frac{1}{9k^2-1}=\frac{1}{2}-\frac{\pi}{6 \sqrt{3}}}(11)

Plugging (11) in (10)

\sum_{k=-\infty}^{\infty}\frac{1}{9k^2-1}=-1+2\left(\frac{1}{2}-\frac{\pi}{6 \sqrt{3}} \right)


\boxed{\sum_{k=-\infty}^{\infty}\frac{1}{9k^2-1}=-\frac{\pi}{3 \sqrt{3}}}(12)

Now, if we Differentiate (9) w.r. to x

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


\boxed{\sum_{k=1}^{\infty}\frac{1}{\left(k^2-x^2\right)^2}=-\frac{1}{2x^4}+\frac{\pi }{4}\left(\frac{\pi x \csc^2\left(\pi x \right)+\cot\left(\pi x \right)}{x^3}\right)}(13)


\text{With}\,\, (13)\,\,\, \text{we can proof}\,\,\, (3), (4),(5) \,\,\text{and}\,\, (6)

\sum_{k=-\infty}^{\infty}\frac{1}{\left(4k^2-1\right)^2}=\sum_{k=-\infty}^{-1}\frac{1}{\left(4k^2-1\right)^2}+1+\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^2}

Letting k=-m in the first sum on the R.H.S., we get

\sum_{k=-\infty}^{\infty}\frac{1}{\left(4k^2-1\right)^2}=+1+2\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^2}(14)

Then

\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^2}=\frac{1}{16}\sum_{k=1}^{\infty}\frac{1}{\left(k^2-\frac{1}{4}\right)^2}

set x=\frac{1}{2} in  (13)

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


\boxed{\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^2}=\frac{\pi^2}{16}-\frac{1}{2}}(15)

Plugging (15) in (14)

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

The proof for (5) and (6) is similar

\sum_{k=-\infty}^{\infty}\frac{1}{\left(9k^2-1\right)^2}=1+2\sum_{k=1}^{\infty}\frac{1}{\left(9k^2-1\right)^2}(17)

Then

\sum_{k=1}^{\infty}\frac{1}{\left(9k^2-1\right)^2}=\frac{1}{81}\sum_{k=1}^{\infty}\frac{1}{\left(k^2-\frac{1}{9}\right)^2}

Setting x=\frac{1}{3} in (13)

\sum_{k=1}^{\infty}\frac{1}{\left(9k^2-1\right)^2}=\frac{1}{81}\left(-\frac{81}{2}+\frac{9 \pi^2 }{4} \csc^2\left(\frac{\pi}{3} \right)+\frac{27 \pi}{4}\cot\left(\frac{\pi}{3} \right) \right)

\sum_{k=1}^{\infty}\frac{1}{\left(9k^2-1\right)^2}=-\frac{1}{2}+\frac{ \pi^2 }{36} \csc^2\left(\frac{\pi}{3} \right)+\frac{ \pi}{12}\cot\left(\frac{\pi}{3} \right) \right)

\sum_{k=1}^{\infty}\frac{1}{\left(9k^2-1\right)^2}=-\frac{1}{2}+\frac{ \pi^2 }{36} \left(\frac{3}{4} \right)+\frac{ \pi}{12}\left(\frac{1}{\sqrt{3}} \right) \right)


\boxed{\sum_{k=1}^{\infty}\frac{1}{\left(9k^2-1\right)^2}=-\frac{1}{2}+\frac{ \pi^2 }{48} +\frac{ \pi}{12\sqrt{3}}}(18)

Plugging (18) in (17)


\boxed{\sum_{k=-\infty}^{\infty}\frac{1}{\left(9k^2-1\right)^2}=\frac{ \pi^2 }{24} +\frac{ \pi}{6\sqrt{3}}}(19)

Now, differentiating (13) w.r. to x

4x\sum_{k=1}^{\infty}\frac{1}{\left(k^2-x^2\right)^3}=\frac{2}{x^5}+\frac{\pi^2}{4}\left(\frac{-2\pi x^2 \csc^2\left(\pi x \right)\cot\left(\pi x \right)-2x\csc^2\left(\pi x \right)}{x^4}\right)+\frac{\pi }{4}\left(\frac{-x^3\pi\csc^2\left(\pi x \right)-3x\cot\left(\pi x \right)}{x^6}}\right)

\sum_{k=1}^{\infty}\frac{1}{\left(k^2-x^2\right)^3}=\frac{1}{2x^6}+\frac{\pi^2}{16}\left(\frac{-2\pi x^2 \csc^2\left(\pi x \right)\cot\left(\pi x \right)-2x\csc^2\left(\pi x \right)}{x^5}\right)+\frac{\pi }{16}\left(\frac{-x^3\pi\csc^2\left(\pi x \right)-3x\cot\left(\pi x \right)}{x^7}}\right)

\sum_{k=1}^{\infty}\frac{1}{\left(k^2-x^2\right)^3}=\frac{1}{2x^6}-\frac{\pi^2 \csc^2\left(\pi x \right)}{8}\left(\frac{\pi x \cot\left(\pi x \right)+1}{x^4}\right)-\frac{\pi }{16}\left(\frac{x^2\pi\csc^2\left(\pi x \right)+3\cot\left(\pi x \right)}{x^6}} \right)(20)

\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^3}=-1+2\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^3}(21)

Then

\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^3}=\frac{1}{64}\sum_{k=1}^{\infty}\frac{1}{\left(k^2-\frac{1}{4}\right)^3}

Letting  x=\frac{1}{2}  in (20), and considering the fact that  \cot\left(\frac{\pi }{2}\right)=0\,\,\,\text{and}\,\,\,\csc^2\left(\frac{\pi }{2}\right)=1    we obtain


\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^3}=\frac{1}{64}\left(\frac{64}{2}-2\pi^2-\pi^2 \right)


\boxed{\sum_{k=1}^{\infty}\frac{1}{\left(4k^2-1\right)^3}=\frac{1}{2}-\frac{3\pi^2}{64}}(22)

Plugging (22) in (21) we get (8)


\boxed{\sum_{k=-\infty}^{\infty}\frac{1}{\left(4k^2-1\right)^3}=-\frac{3\pi^2}{32}}(23)

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