A series

[Grigorios Kostakos]

Prove that:

$$\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({\frac{1}{x+n}+\frac{1}{x+\frac{1}{2}+n}-\frac{2}{2x+n}}\biggr)=2\,\log2-\frac{2}{2x+1}\,,$$

or ,equivenantly, that:

$$\frac{1}{2}\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({\frac{1}{x+n}+\frac{1}{x+\frac{1}{2}+n}-\frac{2}{2x+n+1}}\biggr)=\log2\,.$$

[Tolaso J Kos]

Prove that:

$$\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({\frac{1}{x+n}+\frac{1}{x+\frac{1}{2}+n}-\frac{2}{2x+n}}\biggr)=2\,\log2-\frac{2}{2x+1}\,,$$

or ,equivenantly, that:

$$\frac{1}{2}\displaystyle\mathop{\sum}\limits_{n=1}^{+\infty}\biggl({\frac{1}{x+n}+\frac{1}{x+\frac{1}{2}+n}-\frac{2}{2x+n+1}}\biggr)=\log2\,.$$

We are using the definition of the digamma here stating that:

$$\gamma+\psi(x)=\sum_{n=0}^{\infty}\left ( \frac{1}{n+1}-\frac{1}{x+n} \right )$$

along with the formulas:

\( (1) \quad \displaystyle \sum_{n=0}^{\infty}\frac{1}{n+x}=-\psi(x) \)
\( (2) \quad \displaystyle \Gamma (2x)=\frac{2^{2x-1}}{\sqrt{\pi}}\Gamma (x)\Gamma \left ( x+\frac{1}{2} \right ) \)

Differentiating \( (2) \) with respect to \( x \) (after we take logs) we get:
$$2\psi(2x)=2\ln 2+\psi(x)+\psi\left ( x+\frac{1}{2} \right )$$

We note that the sum can be expressed in digamma form, that is:

$$\sum_{n=0}^{\infty}\left ( \frac{1}{x+n}+\frac{1}{x+\frac{1}{2}+n}-\frac{2}{2x+n}\right )=2\psi(2x)-\psi(x)-\psi\left ( x+\frac{1}{2} \right )$$

Using the formula we derived above we get that:
$$\sum_{n=0}^{\infty}\left ( \frac{1}{x+n}+\frac{1}{x+\frac{1}{2}+n}-\frac{2}{2n+x}\right )=2\ln 2 \Leftrightarrow \sum_{n=1}^{\infty}\left ( \frac{1}{x+n}+\frac{1}{x+\frac{1}{2}+n}-\frac{2}{2n+x}\right )=2\ln 2-\frac{2}{2x+1}$$