INFINITE SUM FROM GRADSHTEYN AND RYZHIK II

     Following the previous post, we will evaluate today the following two infinite sums from Gradshteyn and Ryzhik Table of Integrals, Series, and Products entries 0.238 1 and 2


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


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

We will rely on the the following formulas to compute these two sums:

\psi\left(x+1\right)=-\gamma+\sum_{k=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+x} \right)(1)

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

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

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

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

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

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

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

Now we can use (1) and (2) to rewrite the above equation as

S=-\frac{1}{2}\left(\psi\left(\frac{1}{2}\right)+\gamma\right)-\frac{1}{4}\left(2-\pi \cot\left(\frac{\pi}{2} \right) \right)

S=-\frac{1}{2}\left(-\gamma-2 \ln 2+\gamma\right)-\frac{1}{2}


\boxed{\sum_{k=1}^{\infty}\frac{1}{(2k-1)2k(2k+1)}=\ln 2 - \frac{1}{2}}

Or

\boxed{\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k(4k^2-1)}=\ln 2 - \frac{1}{2}}

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

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

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

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

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

Setting  x=\frac{1}{2}  in (3) we can rewrite

=\sum_{k=1}^{\infty}(-1)^{k-1}\int_0^{1}x^{2k-2}dx-\frac{\ln 2}{2}+\frac{1}{4}\left(2-\pi \csc\left(\frac{\pi}{2} \right)\right)

=\int_0^1\sum_{k=1}^{\infty}(-x^2)^{k-1}dx-\frac{\ln 2}{2}+\frac{1}{2}-\frac{\pi}{4}

=\int_0^1\frac{1}{1+x^2}dx-\frac{\ln 2}{2}+\frac{1}{2}-\frac{\pi}{4}

=\frac{\pi}{4}-\frac{\ln 2}{2}+\frac{1}{2}-\frac{\pi}{4}


\boxed{\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(2k-1)2k(2k+1)}= \frac{1}{2}-\frac{\ln 2}{2}}

Or

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

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