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How to find sum of $\sum_{n=1}^{\infty }\frac{(n)\bmod(k)}{n(n+1)}$?

I know that if $n \bmod k \le k-1$ then this sum is converge then it has finite sum, I just guess it's $\ln(k)$ because when $k=1$ sum is $0=\ln(1)$. I really don't know how to find it. Please help me.

Gary
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HOANXA
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  • Can you do it when $k=2$? – saulspatz May 10 '20 at 14:47
  • I can not now,but it's $ln(2)$ https://www.wolframalpha.com/input/?i=%5Csum+_%7Bn%3D1%7D%5E%7B%5Cinfty+%7D%5C%3A%5Cfrac%7B%5Cleft%28-1%5Cright%29%5E%7Bn%2B1%7D%7D%7Bn%7D – HOANXA May 10 '20 at 15:58
  • Use Taylor series, it's true when $k=2$, $ln(1+x)=1/1-1/2+1/3-1/4...$ when $x=2$ it's $ln(2)$ – HOANXA May 10 '20 at 16:03
  • The real challenge is $n=3$, not $n=2$ – Momo May 10 '20 at 16:24
  • This challenge so .... – HOANXA May 10 '20 at 16:30
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    @HOANXA May I ask... where did you get this problem? – Momo May 10 '20 at 16:42
  • Someones ask me and I really enjoy it, may be he created this ... – HOANXA May 10 '20 at 16:45
  • You should add the example when $k=2$ including an indication of the calculation, to the question body. It's always a good idea to put everything you know about the problem in the question body. – saulspatz May 10 '20 at 16:55
  • @saulspatz That is more for homework-type problems. This is not exactly a homework problem and I don't see a need to add irrelevant content. But I would have liked to know the source of the problem, not just "a friend told me". – Momo May 10 '20 at 18:27
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    @Momo No, it's not for homework type problems only. If you are asking for people to assist you, you should do whatever you can to assist them. This is not a matter of courtesy, or politeness, but simple civility! – saulspatz May 10 '20 at 18:47
  • See the answers from Gary and me. – xpaul May 10 '20 at 19:37
  • @saulspatz 90% of "what user tried" is useless, just put there because moderators forced them. Even here, what n=2 would have helped with the general case? – Momo May 10 '20 at 19:55

5 Answers5

6

Continuing xpaul's answer \begin{align*} & \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\left[ {\psi ^{(0)} \left( {\frac{{r + 1}}{k}} \right) - \psi ^{(0)} \left( {\frac{r}{k}} \right)} \right]} = \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\psi ^{(0)} \left( {\frac{{r + 1}}{k}} \right)} - \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\psi ^{(0)} \left( {\frac{r}{k}} \right)} \\ & = \sum\limits_{r = 1}^{k - 1} {\frac{{r + 1}}{k}\psi ^{(0)} \left( {\frac{{r + 1}}{k}} \right)} - \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\psi ^{(0)} \left( {\frac{r}{k}} \right)} - \sum\limits_{r = 1}^{k - 1} {\frac{1}{k}\psi ^{(0)} \left( {\frac{{r + 1}}{k}} \right)} \\ & = \sum\limits_{r = 2}^k {\frac{r}{k}\psi ^{(0)} \left( {\frac{r}{k}} \right)} - \sum\limits_{r = 1}^{k - 1} {\frac{r}{k}\psi ^{(0)} \left( {\frac{r}{k}} \right)} - \frac{1}{k}\sum\limits_{r = 2}^k {\psi ^{(0)} \left( {\frac{r}{k}} \right)} \\ & = \psi ^{(0)} (1) - \frac{1}{k}\sum\limits_{r = 1}^k {\psi ^{(0)} \left( {\frac{r}{k}} \right)} = - \gamma + \gamma + \log k = \log k. \end{align*}

Gary
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    That's great, upvote+1). – xpaul May 10 '20 at 19:22
  • @ Gary Could you please explain the last line, i.e. how did you calculate the sum over the digamma function? BTW because I did not understand it I developed my solution with the integral method. – Dr. Wolfgang Hintze Mar 04 '21 at 20:20
  • @Dr.WolfgangHintze I just looked it up. It is the first one in https://en.wikipedia.org/wiki/Digamma_function#Some_finite_sums_involving_the_digamma_function I did not care to prove it myself. Thanks for the alternative solution. – Gary Mar 05 '21 at 06:32
  • @ Gary Thanks. You might wish to indicate this reference in your answer, in particular because it is due to the great Gauß. A proof would be even better. – Dr. Wolfgang Hintze Mar 05 '21 at 17:39
5

Note that $n=mk+r$, $r=0,1,\cdots k-1$. So \begin{eqnarray} &&\sum_{n=1}^{\infty }\frac{(n)\mod(k)}{n(n+1)}=\sum_{m=0}^{\infty }\sum_{r=1}^{k-1}\frac{(mk+r)\mod(k)}{(mk+r)(mk+r+1)}\\ &=&\sum_{m=0}^{\infty }\sum_{r=1}^{k-1}\frac{r}{(mk+r)(mk+r+1)}=\sum_{r=1}^{k-1}r\sum_{m=0}^{\infty }\frac{1}{(mk+r)(mk+r+1)}\\ &=&\sum_{r=1}^{k-1}r\sum_{m=0}^{\infty }\bigg(\frac{1}{mk+r}-\frac{1}{mk+r+1}\bigg)=\sum_{r=1}^{k-1}\frac{r}{k}\sum_{m=0}^{\infty }\bigg(\frac{1}{m+\frac{r}{k}}-\frac{1}{m+\frac{r+1}{k}}\bigg)\\ &=&\sum_{r=1}^{k-1}\frac{r}{k}\sum_{m=1}^{\infty }\bigg(\frac{1}{m+\frac{r}{k}-1}-\frac{1}{m+\frac{r+1}{k}-1}\bigg)\\ &=&\sum_{r=1}^{k-1}\frac{r}{k}\sum_{m=1}^{\infty }\bigg[\bigg(\frac1m-\frac{1}{m+\frac{r+1}{k}-1}\bigg)-\bigg(\frac{1}{m}-\frac{1}{m+\frac{r}{k}-1}\bigg)\bigg]\\ &=&\sum_{r=1}^{k-1}\frac{r}{k}\bigg[\psi ^{(0)}\left(\frac{r+1}{k}\right)-\psi ^{(0)}\left(\frac{r}{k}\right)\bigg]. \end{eqnarray} Here the Digamma function $$ \psi^{(0)}(z+1)=-\gamma+\sum_{n=1}^\infty\bigg(\frac1n-\frac{1}{n+z}\bigg) $$ is used from here.

Update: Now @Gary proves $$ \sum_{r=1}^{k-1}\frac{r}{k}\bigg[\psi ^{(0)}\left(\frac{r+1}{k}\right)-\psi ^{(0)}\left(\frac{r}{k}\right)\bigg]=\ln k. $$ Thank you, Gary.

xpaul
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Also, @xpaul 's computation shows that the initial sum is asymptotically equal to $\log k$. Indeed, @xpaul derived \begin{eqnarray} &&\sum_{n=1}^{\infty }\frac{(n)\mod(k)}{n(n+1)}=\sum_{m=0}^{\infty }\sum_{r=1}^{k-1}\frac{(mk+r)\mod(k)}{(mk+r)(mk+r+1)}\\ &=&\sum_{m=0}^{\infty }\sum_{r=1}^{k-1}\frac{r}{(mk+r)(mk+r+1)}=\sum_{r=1}^{k-1}r\sum_{m=0}^{\infty }\frac{1}{(mk+r)(mk+r+1)} \end{eqnarray} and we can estimate the inner sum as follows $$ \frac{1}{r(r+1)}<\sum_{m=0}^{\infty }\frac{1}{(mk+r)(mk+r+1)}< \frac{1}{r(r+1)} + \sum_{m=1}^{\infty }\frac{1}{(mk)^2}< \frac{1}{r(r+1)}+\frac{10}{k^2}. $$ Plugging this to the previous expression we get \begin{align*} \sum_{n=1}^{\infty }\frac{(n)\mod(k)}{n(n+1)}=\sum_{r=1}^{k-1}\frac{1}{r+1}+\mathcal{O}(k^{-1})=\log k +\mathcal{O}(1) \end{align*}

Tony419
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Using the same ideas as in a recent answer (https://math.stackexchange.com/a/4024522/198592) here is a solution which is very close to that of @xpaul and @Gary but more elementary in the final step (without polygamma functions).

We will prove that

$$s(k):=\sum _{n=1}^{\infty } \frac{n \bmod k}{n (n+1)}=\log (k)\tag{1}$$

  1. simplify the mod function letting

$$n = m k +j, m=0,1,2,..., j=1, 2, ..., k-1\tag{2}$$

notice that $j$ runs only up to $k-1$ because $(m k + k) mod(k) = 0$.

Then $n \bmod k = j$ and defining

$$u(k,j) = \sum_{m=0}^{\infty} \frac{j}{(m k+j)(m k+j+1)}\tag{3}$$

we get

$$s(k) = \sum_{j=1}^{k-1} u(k,j)\tag{4}$$

  1. Calculate $u$

Instead of doing the $m$-sum directly which leads to polygamma functions we prefer to represent denominators by integrals using the formula $\frac{1}{a}=\int_{0}^{\infty} e^{-a t}\,dt$

After partial fraction decomposition

$$\frac{j}{(j+k m) (j+k m+1)}=\frac{j}{j+k m}-\frac{j}{j+k m+1}$$

this gives after also doing the $m$-sum

$$u(k,j) = \int_{0}^{\infty} j\sum_{m=0}^{\infty}\left(e^{-t(j+m k)}-e^{-t(1+j+m k)} \right)\,dt=\int_{0}^{\infty} j\left( \frac{e^{k t-j t}}{e^{k t}-1}-\frac{e^{k t-(j+1) t}}{e^{k t}-1}\right)\,dt\tag{5}$$

  1. do the $j$-sum under the integral gives for the integrand

$$\sum_{j=1}^{k-1} j\left( \frac{e^{k t-j t}}{e^{k t}-1}-\frac{e^{k t-(j+1) t}}{e^{k t}-1}\right) = \frac{k \left(-e^t\right)+e^{k t}+k-1}{\left(e^t-1\right) \left(e^{k t}-1\right)}=\frac{1}{e^t-1}-\frac{k}{e^{k t}-1}\tag{6}$$

  1. the final step is to calculate the $t$-integral

$$s(k) = i(k) := \int_{0}^{\infty} \left(\frac{1}{e^t-1}-\frac{k}{e^{k t}-1}\right)\,dt\tag{7}$$

The indefinte integral (antderivative) is

$$a(t) = \int \left(\frac{1}{e^t-1}-\frac{k}{e^{k t}-1}\right)\,dt\\ =-t+k t+\log \left(\frac{e^t-1}{e^{k t}-1}\right)\tag{8}$$

and for the limits at the integration boundaries we find

$$a(t \to \infty) = 0, a(t \to 0) = \log(\frac{1}{k})\tag{9}$$

Hence the integral $(7)$ becomes $i(k)=\log(k)$ which completes the proof.

Details of the limits

$$\begin{align} a(t \to 0) : & -t + k t+\log \left(\frac{e^t-1}{e^{k t}-1}\right)\\ & \to \log \left(\frac{\left(1+t+...\right)-1}{\left(1+k t +...\right)-1}\right)\\ & \to \log \left(\frac{1}{k}\right) \end{align}\tag{10a}$$

$$\begin{align} a(t \to \infty) : & -t + k t+\log \left(\frac{e^t-1}{e^{k t}-1}\right)\\ & \to -t +k t+\log \left(\frac{e^t \left(1-e^{-t}\right)}{e^{k t} \left(1-e^{-k t}\right)}\right)\\ & \to -t+k t+\log \left(\frac{e^t}{e^{k t}}\right)+\log \left(\frac{1-e^{-t}}{1-e^{-k t}}\right)\\ & = \log \left(\frac{1-e^{-t}}{1-e^{-k t}}\right) \to 0 \end{align}\tag{10b}$$

Discussion

The integral $i(k)$ extends the domain of $s(k)$ to real values of $k$.

Dr. Wolfgang Hintze
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  • Testing the integration method with the sum $s_1(k ) = \sum_{n=1}^{\infty} \frac{n ; mod(k)}{n^2}$ I got stuck at a tough integral ($\int \frac{k t}{\left(e^t-1\right) \left(e^{k t}-1\right)} , dt$), and I conclude temporarily that the finite sum of polygamma functions (which appear by doing the $m$-sum) cannot be further reduced in general, and that the integral method is restricted to specific cases. But, reversing the logic, the tough integral was shown to be the finite sum over polygamma functions. Which is nice to know. – Dr. Wolfgang Hintze Mar 04 '21 at 16:35
  • can you show me the detail of (5), why that series equal to $\left( \frac{e^{k t-j t}}{e^{k t}-1}-\frac{e^{k t-(j+1) t}}{e^{k t}-1}\right)$ ? – HOANXA Mar 04 '21 at 17:13
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    It just summing geometric sums. Can you see it now? – Dr. Wolfgang Hintze Mar 04 '21 at 18:46
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My idea is:

$\sum \frac{nmodk}{n(n+1)}=\frac{1}{1.2}+\frac{2}{2.3}+...+\frac{k-1}{(k-1)k}+...=\frac{1}{1}-\frac{1}{2}+\frac{2}{2}-\frac{2}{3}+...$ $=(\frac{\:1}{1\:}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}+\frac{k-1}{k})+(\frac{\:1}{k+1\:}+\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k-1}+\frac{k-1}{2k})+...$ $=\sum _{n=1}^{\infty }\left(\frac{1}{\left(n-1\right)k+1}+\frac{1}{\left(n-1\right)k+2}+...+\frac{1}{\left(n-1\right)k+k-1}-\frac{k-1}{\left(n-1\right)k+k}\right)$

But it still not work

HOANXA
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