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.
 A: Let $m\geq 2$. Put:
$$S_m=\sum_{n\geq 0}(\frac{1}{mn+1}+\cdots+\frac{1}{mn+m-1}-\frac{m-1}{mn+m}) $$
As $\displaystyle \frac{1}{mn+r}=\int_0^1x^{mn+r-1}dx$, We have 
$$S_m=\int_0^{1}\frac{1+x+\cdots+x^{m-2}-(m-1)x^{m-1}}{1-x^m} dx$$
But
$$1+x+\cdots+x^{m-2}-(m-1)x^{m-1}=(1-x^{m-1})+\cdots+(x^{m-2}-x^{m-1})$$
Hence 
$$1+x+\cdots+x^{m-2}-(m-1)x^{m-1}=[(1-x)(1+x+\cdots x^{m-2})]+\cdots+[x^{m-2}(1-x)]$$
and
$$[(1+x+\cdots x^{m-2})]+\cdots+[x^{m-2}]=1+2x+\cdots+(m-1)x^{m-2}$$
Thus
$$S_m=\int_0^1\frac{\sum_{k=0}^{m-2}(k+1)x^{k}}{1+x+\cdots+x^{m-1}}dx=\int_0^1\frac{P^{\prime}(x)}{P(x)}dx=[\log(1+x+\cdots+x^{m-1})]_0^1=\log(m)$$
A: 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.
A: So far, there is a proof involving differentiation/integration, and a proof involving nothing more than the knowledge that $\gamma$amma exists. Here is my proof, requiring the Taylor series expansion of the logarithm.
The well known Taylor series for $\ln(x)$ is as follows:
$$-\ln(1-x)=\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\cdots+\frac{x^m}{m}+\cdots$$
which I love, because the exponents correspond to the denominators.
Substituting in $x^m$, we get:
$$-\ln(1-x^m)=\frac{x^m}{1}+\frac{x^{2m}}{2}+\frac{x^{3m}}{3}+\cdots=\frac{m\,x^m}{m}+\frac{m\,x^{2m}}{2m}+\frac{m\,x^{3m}}{3m}+\cdots$$
where I rewrote it slightly so that the exponents still correspond to the denominators.
Subtracting the bottom from the top, we get:
$$\ln\left(\frac{1-x^m}{1-x}\right)=\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}+\cdots+\frac{-(m-1)x^m}{m}+\\
\frac{x^{m+1}}{m+1}+\cdots+\frac{-(m-1)x^{2m}}{2m}+\cdots$$
(i.e. the coefficients are $-(m-1)$ where the denominator is a multiple of $m$, and $1$ otherwise.)
The LHS can be rewritten as $\ln(1+x+x^2+\cdots+x^m)$. So, plugging in $x=1$:
$$\ln(m)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots-\frac{m-1}{m}+\frac{1}{m+1}+\cdots-\frac{-(m-1)}{2m}+\cdots$$
which is what was meant to be proved.
A: The partial sums of this series can be written in a simple form. The generic term is
$$
\frac1{mk+1}+\cdots+\frac1{mk+(m-1)}-\frac{m-1}{mk+m}=\left(\sum_{j=1}^m\frac1{mk+j}\right)-\frac1{k+1},$$
so the sum of the first $n$ terms is
$$\sum_{k=0}^{n-1}\left(\sum_{j=1}^m\frac1{mk+j}\right)-\sum_{k=0}^{n-1}\frac1{k+1}=\sum_{k=1}^{mn}\frac1k-\sum_{k=1}^n\frac1k=\sum_{k=n+1}^{mn}\frac1k.$$
Let $n\to\infty$. The limit of the RHS is $\log(m)$, by this result:

Claim: Let $(a_n)$ and $(b_n)$ be sequences of positive integers with $b_n\ge a_n\to\infty$ and $\lim_{n\to\infty}\frac{b_n}{a_n}=c$. Then
$$
\lim_{n\to\infty}\sum_{k=a_n}^{b_n}\frac1k=\log c.$$
A: Having $\lim_{n\rightarrow\infty}$ for all expressions, 
(especially for $\left(H_{xn}-H_n\right)$ combinations to avoid $\infty-\infty$ complications) 
and 
having $H_n=\sum_{1}^n\frac 1i$;
No 1 expression left side equals: $H_{2n}-H_n$
No 2 expression left side equals: $H_{3n}−H_n$
No 3 expression left side equals: $H_{4n}−H_n$
Having  as Euler–Mascheroni constant;
$$=H_n-\ln n$$
Generalizing:
$$=H_{xn}-\ln xn$$
$$\Rightarrow H_{xn}-H_n=\ln xn -\ln n$$
$$\Rightarrow H_{xn}-H_n=\ln x $$
where $x\geq 1$.
Accordingly;
No 1 expression: $H_{2n}-H_n=\ln 2$
No 2 expression: $H_{3n}−H_n=\ln 3$
No 3 expression: $H_{4n}−H_n=\ln 4$
