I would like to find the asymptotics of


I saw Find asymptotics of $\sum\limits_{i=0}^{n/3} 2^i \binom{n-i-1}{\frac{2n}{3}-1}$ but I can't get anything similar to work in my example.

What is the right way to do this?

  • $\begingroup$ f[n_] := Sum[Binomial[n, i]/i^((n + 1)/2), {i, 1, n}]; a = Table[f[n], {n, 1, 100}] $\endgroup$ Dec 18, 2013 at 20:52
  • $\begingroup$ @belisarius This seems to imply the asymptotic is exactly $n$. $\endgroup$
    – marshall
    Dec 18, 2013 at 20:54
  • $\begingroup$ Now prove it :) $\endgroup$ Dec 18, 2013 at 21:07

1 Answer 1


Let's start by looking at this graph

 Sum[Binomial[n, i]/i^((n + 1)/2), {i, 1, n}]/n, {n, 500, 550}]

enter image description here

So we have the hypothesis that the sum increases like $n$. Let's use Mathematica to prove that the sum $S$ divided by $n$ goes to 1.

We have

Limit[1/n Binomial[n, i]/i^n, n -> Infinity, Assumptions -> i >= 2]


So all but the first term go to 0. But we have a lot of such terms, so this is not enough to say the "rest of the sum" $R$ goes to 0. We prove that the sum from i=3 onwards goes to 0.

We have

Limit[n Binomial[n, n/2]/3^n, n -> Infinity]


This proves that the sum from i=3 onwards goes to 0, as the Binomial coefficient is maximal when i=n/2, 1/(i^n) is maximal when i=3 and we have n-2 ~ n terms.

The second term also goes to 0, so indeed the "rest of the sum", i.e. $R$, goes to 0.

The first term is

n == Binomial[n,1]*1^x

So we have that $$\lim_{n\rightarrow \infty} \frac{S}{n} = \lim_{n\rightarrow \infty} \frac{n}{n} + \lim_{n\rightarrow \infty} \frac{R}{n} = 1+0 = 1$$


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