<s>[INST] Series for logarithms This is more of a challenge than a question, but I thought I'd share anyway. Prove the following identities, and prove that the pattern continues.
\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)=\ln2
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{3n+1}+\frac{1}{3n+2}-\frac{2}{3n+3}\right)=\ln3
\end{equation*}\begin{equation*}
\sum_{n=0}^\infty\left(\frac{1}{4n+1}+\frac{1}{4n+2}+\frac{1}{4n+3}-\frac{3}{4n+4}\right)=\ln4
\end{equation*}\begin{equation*}
\mathrm{etc.}
\end{equation*}
By the way, a good notation for discussing this problem can be found here. The problem is to prove that:
$[\overline{1,-1}]=\ln2,\;\;$ $[\overline{1,1,-2}]=\ln3,\;\;$ $[\overline{1,1,1,-3}]=\ln4,\;\;$ $[\overline{1,1,1,1,-4}]=\ln5,\;\;$ etc.
 [/INST] For any $m \in \mathbb{N},$
$$\sum_{k=0}^{\infty} \Big(\frac{1}{mk+1} + \frac{1}{mk+2} + ... + \frac{1}{mk+m-1}- \frac{m-1}{mk+m}\Big)$$ $$= \lim_{n \rightarrow \infty} \Big(\sum_{k=0}^{mn} \frac{1}{mk+1} + ... + \frac{1}{mk+m-1} - \frac{m-1}{mk+m}\Big)$$$$= \lim_{n \rightarrow \infty} \Big( \sum_{k=1}^{mn} \frac{1}{k} - m \sum_{k=1}^n \frac{1}{mk} \Big)$$ $$= \lim_{n \rightarrow \infty} \Big(\sum_{k=1}^{mn} \frac{1}{k} - \sum_{k=1}^n \frac{1}{k}\Big)$$ $$= \lim_{n \rightarrow \infty} \Big( \sum_{k=1}^{mn} \frac{1}{k} - \ln(mn) + \ln(mn) - \sum_{k=1}^n \frac{1}{k} \Big)$$$$= \lim_{n \rightarrow \infty} \Big( \sum_{k=1}^{mn} \frac{1}{k} - \ln(mn) + \ln(n) - \sum_{k=1}^n \frac{1}{k} \Big) + \ln(m)$$ $$= \gamma - \gamma + \ln(m) = \ln(m),$$ where $\gamma$ is the Euler-Mascheroni constant.
</s>