On the sums $\sum\limits_{i=0}^n\frac{i}{n}\ln(\frac{i}{n})$ and $\sum\limits_{i=0}^n(-1)^i\frac{i}{n}\ln(\frac{i}{n})$

I was thinking about unimodal sequences, and the two which occurred to me are $$\binom{n}{i}$$ and $$\dfrac{i}{n}\ln(\dfrac{i}{n})$$, both for $$i=0$$ to $$n$$ (for the second, its value is $$0$$ at $$i=0$$).

For the first, it is well known that $$\sum_{i=0}^n \binom{n}{i} =2^n$$ and $$\sum_{i=0}^n (-1)^i\binom{n}{i} =0$$.

I naturally wondered about the corresponding results for $$A_n =\sum_{i=0}^n\dfrac{i}{n}\ln(\dfrac{i}{n})$$ and $$A_n^{\pm} =\sum_{i=0}^n(-1)^i\dfrac{i}{n}\ln(\dfrac{i}{n})$$.

Here's what I have shown.

$$A_n = -\dfrac{n}{4}+\dfrac{\ln(n)}{12n}+\dfrac1{4n}+O\left(\dfrac1{n^2}\right)$$ $$A_{2n}^{\pm} =\dfrac{3\ln(n)}{8n}+O\left(\dfrac1{n}\right)$$ $$A_{2n+1}^{\pm} =\dfrac{\ln(n)}{8n}+O\left(\dfrac1{n}\right)$$

I have verified these computationally.

My proofs, as they often are, are fairly messy, especially for $$A_{n}^{\pm}$$, so my questions are (ya gotta have a question)

1. How well known are these results?
2. Are there reasonably simple proofs of them?
3. Is there a simple proof that $$A_{n}^{\pm} \to 0$$ as $$n \to \infty$$?
• are you sure your sums start at $i=0$? Jan 15, 2020 at 5:55
• You can start them at 1 if it makes you feel better. I consider them a discrete version (for $A_n$) of $\int_0^1 x\ln(x) dx$, so the terms are zero at $i=0$. Jan 15, 2020 at 5:59
• The convention $0\log 0=0$ is often used. Jan 15, 2020 at 6:01
• For the first one, if started from $i=1$, the Mathematica gives a longish expression in terms of Zeta and Log(Glaisher)! Jan 15, 2020 at 6:24

As @Dr Zafar Ahmed DSc commented

$$A_n=\sum_{i=1}^n\left(\frac{i}{n}\right)\log \left(\frac{i}{n}\right)=\frac{12 \zeta ^{(1,0)}(-1,n+1)+12 \log (A)-6 n(n+1) \log \left({n}\right)-1}{12 n}$$

For large values of $$n$$ $$A_n=-\frac{n}{4}+\frac{12\log (A)+ \log \left({n}\right)}{12n}+\frac{1}{720 n^3}+O\left(\frac{1}{n^5}\right)$$

Computing $$A_n+\frac{n}{4}$$ with $$n=10^k$$ $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 1 & 0.044065045700551029177 & 0.044065043726241327229 \\ 2 & 0.006325187980883474321 & 0.006325187980863634043 \\ 3 & 0.000824400751671184572 & 0.000824400751671184374 \\ 4 & 0.000101628284137902171 & 0.000101628284137902171 \\ 5 & 0.000012081649324481089 & 0.000012081649324481089 \end{array} \right)$$

We also have $$\frac{A_{n+1}}{A_n}=1+\frac{1}{n}+\frac{24 \log (A)+2 \log (n)-1}{3 n^3}+O\left(\frac{1}{n^4}\right)$$

• I found my error - my 6n should be 12n as shown here. My 1/n term still differs, but I don't want to chase that down. Jan 15, 2020 at 23:28
• @martycohen. No problem with the $\frac 1n$ term since $\log(A)=0.248754\sim \frac 14$ !! Jan 16, 2020 at 3:33
• But that's for small values of $\frac14$. Jan 16, 2020 at 4:05
• Also see this related question of mine: math.stackexchange.com/questions/3510666/… Jan 16, 2020 at 4:06