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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 $n \to \infty$?
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    $\begingroup$ are you sure your sums start at $i=0$? $\endgroup$
    – clathratus
    Jan 15, 2020 at 5:55
  • $\begingroup$ 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 $i=0$. $\endgroup$ Jan 15, 2020 at 5:59
  • $\begingroup$ The convention $0\log 0=0$ is often used. $\endgroup$ Jan 15, 2020 at 6:01
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    $\begingroup$ For the first one, if started from $i=1$, the Mathematica gives a longish expression in terms of Zeta and Log(Glaisher)! $\endgroup$
    – Z Ahmed
    Jan 15, 2020 at 6:24

1 Answer 1

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

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  • $\begingroup$ 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. $\endgroup$ Jan 15, 2020 at 23:28
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    $\begingroup$ @martycohen. No problem with the $\frac 1n$ term since $\log(A)=0.248754\sim \frac 14$ !! $\endgroup$ Jan 16, 2020 at 3:33
  • $\begingroup$ But that's for small values of $\frac14$. $\endgroup$ Jan 16, 2020 at 4:05
  • $\begingroup$ Also see this related question of mine: math.stackexchange.com/questions/3510666/… $\endgroup$ Jan 16, 2020 at 4:06

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