# 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.

• "challenge" meaning that you already know how to do it? Jul 31, 2014 at 6:02
• Indeed. At the time of writing this comment, one solution has been posted - that way was the way I first did it, too. However, I have since come up with a "nicer" way (without using derivatives or integrals - though it does use the Taylor series for logarithms). Jul 31, 2014 at 6:44
• Apparently, this could also be made computing $$\sum_{n=0}^{\infty}(\frac{1}{mn+i}-\frac{1}{mn+m})=-\frac{\psi ^{(0)}\left(\frac{i}{m}\right)+\gamma }{m}$$ But I am stuck with the summation of the digamma functions. Jul 31, 2014 at 7:06
• Hello! There is a way to do this using no calc; only the fact that $e^x\ge x+1$ and the "intuitively obvious" squeeze theorem. Mar 2, 2015 at 2:31
• math.stackexchange.com/questions/46378/… Apr 21, 2015 at 15:43

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 $$S_m=\int_0^1\frac{\sum_{k=0}^{m-2}(k+1)x^{k}}{1+x+\cdots+x^{m-1}}dx=\int_0^1\frac{P^{\prime}(x)}{P(x)}dx=[\log(1+x+\cdots+x^{m-1})]_0^1=\log(m)$$

• This is a really nice answer ! Thanks for providing this solution. Cheers :) Jul 31, 2014 at 6:34
• This was my first solution, too! However, I have since come up with a "nicer" solution in the shower (it doesn't involve any integrals or derivatives, but it does require knowledge of the Taylor series for ln). Jul 31, 2014 at 6:42
• Could you show this solution ? Jul 31, 2014 at 7:18
• Sure - as soon as I get on my computer (am currently on mobile). $\LaTeX$ and phones do not mix. Jul 31, 2014 at 19:27
• @ClaudeLeibovici Done. Jul 31, 2014 at 23:01

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.

• Wow. That was an interesting approach. Good job! Jul 31, 2014 at 15:59
• Not exactly the same but the second last line in that proof above resembles the answer to this question here: math.stackexchange.com/questions/46713/… Apr 22, 2015 at 11:50

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.

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 $$\lim_{n\to\infty}\sum_{k=a_n}^{b_n}\frac1k=\log c.$$

• Nice! How did you find a 7.5-year-old question? Jan 19, 2022 at 0:44
• Haha, from this even older question math.stackexchange.com/q/46378/215011 which can be solved the same way. Jan 19, 2022 at 0:56
• @AkivaWeinberger The same device can be applied to your question math.stackexchange.com/q/883233/215011 , if you're interested in seeing an elementary solution Jan 20, 2022 at 6:34

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 𝛾 as Euler–Mascheroni constant; $$𝛾=H_n-\ln n$$

Generalizing: $$𝛾=H_{xn}-\ln xn$$ $$\Rightarrow H_{xn}-H_n=\ln xn -\ln n$$ $$\Rightarrow H_{xn}-H_n=\ln x$$ 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: $H_{4n}−H_n=\ln 4$

• This is essentially @user165670's answer above, but with fewer details and less rigour. Nov 15, 2016 at 18:44
• @AkivaWeinberger, to my surprise our comments seem to have vanished! Have you flagged them to a moderator for deletion? I wouldn't mind that, since they were not valuable, but I'd like to know (for my curiosity) how they disappeared. Nov 15, 2016 at 19:47
• @AlexM. …No, I did not… I don't know who deleted them Nov 15, 2016 at 19:52
• I just wanted to emphasize the concept of $H_xn-H_n=\ln x$. I was excited with the idea of $H_{-n]-H_n=\ln {-1}$. Nov 16, 2016 at 22:59