A Question based on series. Find the sum of$$\sum _{r=1}^\infty\left({\frac{2}{4r-3}-\frac{1}{2r}}\right)$$ I tried to solve this problem by taking $2$ common from the expression and got the result as of$$2\sum _{r=1}^\infty\left({\frac{1}{4r-3}-\frac{1}{4r}}\right)$$ but I could not get any constant term on which I could apply the telescopic series formula please help me out.
This is not a telescopic series, though it might superficially resemble one.

The common technique with such problems is to transform them to power series (by putting $x^n$ or something in the numerator), then fiddle around with functions, their Taylor series, maybe some differentiation or other stuff, and then evaluate the series at $x=1$.

Short of that, we are reduced to using the following crude tools.

Suppose you know that $\sum\limits_{n=1}^\infty{(-1)^{n+1}\over n}=\ln2$ (because if you don't, we aren't going anywhere). Suppose you also know that $\sum\limits_{n=1}^\infty{(-1)^{n+1}\over 2n+1}={\pi\over4}$, which is also a mandatory prerequisite. Then, obviously
$$\sum_{n=1}^\infty{(-1)^{n+1}\over 2n}={1\over2}\sum_{n=1}^\infty{(-1)^{n+1}\over n}={\ln2\over2}$$
Now I'm going to put it in a simple way, without all those fancy $\sum$s:
$$\begin{array}{ccccccccccc}
{\pi\over4} &amp;=&amp; 1 &amp; &amp; -{1\over3} &amp; &amp; +{1\over5} &amp; &amp; -{1\over7} &amp;\dots \\
\ln 2 &amp;=&amp; 1 &amp; -{1\over2} &amp; +{1\over3} &amp; -{1\over4} &amp; +{1\over5} &amp; -{1\over6} &amp; +{1\over7} &amp;\dots \\
{\ln 2\over2} &amp;=&amp; &amp; \phantom{-}{1\over2} &amp; &amp; -{1\over4} &amp; &amp; +{1\over6} &amp; &amp;\dots \\
\end{array}$$
Now just add these three series together.
$$S_n=\sum _{r=1}^n \left(\frac{2}{4r-3}-\frac{1}{2r}\right)=\frac12\sum _{r=1}^n \left(\frac{1}{r+\frac14-1}-\frac{1}{r+1-1}\right)$$
It is a easy problem to people familiar with the polygamma functions.
$$\sum _{r=1}^n \frac{1}{r+a-1}=\psi(a+n)-\psi(a)$$
$\psi(x)$ is the digamma function.

Successively with $a=\frac14$ and $a=1$ :
$$S_n=\frac12\left(\psi\left(\frac14+n\right)-\psi\left(\frac14\right)-\psi\left(1+n\right)+\psi\left(1\right)\right)$$
$n\to\infty\quad \psi(a+n)\sim \ln(n)+O\left(\frac{1}{x}\right)$

$\lim_{n\to\infty}\left(\psi\left(\frac14+n\right)-\psi\left(1+n\right)\right)=0$
$$S_\infty=\frac12\left(-\psi\left(-\frac14\right)+\psi\left(1\right)\right)$$
$\psi(1)=-\gamma$ Euler-Mascheroni constant. And $\psi(1/4)=-\gamma-\frac{\pi}{2}-3\ln(2)$
$$S_\infty=\frac12\left(\frac{\pi}{2}+3\ln(2)\right)$$
$$\sum _{r=1}^\infty \left(\frac{2}{4r-3}-\frac{1}{2r}\right)=\frac{\pi}{4}+\frac32\ln(2)$$
For reference : http://mathworld.wolfram.com/DigammaFunction.html