\(\sum_{n=1}^{\infty}\bigl({\frac{1}{n}-\log\bigl({\frac{n+1}{n}}\bigr)}\bigr)\)
- Grigorios Kostakos
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\(\sum_{n=1}^{\infty}\bigl({\frac{1}{n}-\log\bigl({\frac{n+1}{n}}\bigr)}\bigr)\)
Find the sum \[\displaystyle\mathop{\sum}\limits_{n=1}^{\infty}\biggl({\frac{1}{n}-\log\bigl({\tfrac{n+1}{n}}\bigr)}\biggr)\,.\]
Grigorios Kostakos
Re: \(\sum_{n=1}^{\infty}\bigl({\frac{1}{n}-\log\bigl({\frac{n+1}{n}}\bigr)}\bigr)\)
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.
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.
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- Grigorios Kostakos
- Founder
- Posts: 461
- Joined: Mon Nov 09, 2015 1:36 am
- Location: Ioannina, Greece
Re: \(\sum_{n=1}^{\infty}\bigl({\frac{1}{n}-\log\bigl({\frac{n+1}{n}}\bigr)}\bigr)\)
To be precise: \[\mathop{\lim}\limits_{m\to \infty} \biggl(-\log(m+1) + \sum_{n=1}^m \frac{1}{n}\biggr)=\mathop{\lim}\limits_{m\to \infty} \biggl(-\frac{1}{m+1}-\log(m+1) +\sum_{n=1}^{m+1} \frac{1}{n}\biggr)=0+\gamma=\gamma\,.\]
Grigorios Kostakos
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