\(\sum_{n=1}^{\infty}\bigl({\frac{1}{n}-\log\bigl({\frac{n+1}{n}}\bigr)}\bigr)\)

[Grigorios Kostakos]

Find the sum \[\displaystyle\mathop{\sum}\limits_{n=1}^{\infty}\biggl({\frac{1}{n}-\log\bigl({\tfrac{n+1}{n}}\bigr)}\biggr)\,.\]

[admin]

Replied by ex-member aziiri:

First, let us take the sum up to \(m\) instead of \(\infty\).
\[\sum_{n=1}^m \frac{1}{n} - \log\left(\frac{n+1}{n} \right)= \sum_{n=1}^m \frac{1}{n} -\sum_{n=1}^m \bigl(\log(n+1)-\log(n)\bigr).\] Telescoping the last sum gives us that the requested sum is : \[\lim_{m\to \infty} -\log(m+1) + \sum_{n=1}^m \frac{1}{n}.\] Which is the definition of the Euler Mascheroni constant.

[Grigorios Kostakos]

To be precise: \[\mathop{\lim}\limits_{m\to \infty} \biggl(-\log(m+1) + \sum_{n=1}^m \frac{1}{n}\biggr)=\mathop{\lim}\limits_{m\to \infty} \biggl(-\frac{1}{m+1}-\log(m+1) +\sum_{n=1}^{m+1} \frac{1}{n}\biggr)=0+\gamma=\gamma\,.\]