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I'd like to ask an upper bound $C_n$ such that $$ S_n(k):=n^{-k}\sum_{i=1}^{n-1}\binom{n}{i}i^k\leq C_n\cdot(1-1/n)^k $$ for all (sufficiently large) $k$. A naive upper bound is to replace all $i^k$ with $(n-1)^k$, which produces $$ S_n(k)\leq(2^n-2)\cdot(1-1/n)^k. $$ On the other hand, by only considering $i=n-1$ term one get $$ S_n(k)\geq n\cdot(1-1/n)^k. $$ I'd like to ask if one can get $C_n$ better than $2^n-2$ (namely, growth slower as $n\to\infty$)?

I notice that $$ \sum_{i=1}^{n-1}\binom{n}{i}i^k=\left(X\frac{\mathrm d}{\mathrm dX}\right)^k \big((1+X)^n-1-X^n\big)\bigg|_{X=1}. $$ But I failed to find a way to produce an upper bound from this.

Any suggestions?

EDIT: It turns out that I have asked a stupid question. It is equivalent to ask an upper bound of $$ C_n(k):=(n-1)^k\sum_{i=1}^{n-1}\binom{n}{i}i^k =\sum_{i=1}^{n-1}\binom{n}{i}\left(\frac{i}{n-1}\right)^k $$ for either (i) all $k\geq 0$ or (ii) for sufficiently large $k$. It's clear that $C_n(k)$ decreases as $k$ increases. For (i) the $2^n-2$ cannot be improved since $C_n(0)=2^n-2$. For (ii) it's $n+\epsilon$ for any $\epsilon>0$ (here the "sufficiently large" depending on $n$ and $\epsilon$) since $\lim_{k\to\infty}C_n(k)=n$.

What estimation is useful in my application is still not clear to me. Perhaps choose a $k_0$ (independent of $n$) and ask a simple upper bound of $C_n(k_0)$?

Jz Pan
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    Notice your sum is $\sum_{i=0}^n \binom{n}{i} i^k$ minus $n^k$. But this sum is very well studied, and it has a closed form for each $k$ (this is basically because the sum is hypergeometric). Getting this closed form to be uniform in $k$ is hard, and has been asked many times before. See here, here, here, here, etc. – HallaSurvivor May 03 '23 at 18:45
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    Since you're interested in asymptotics, note that the sum $$\sum_{i=0}^n \binom{n}{i} i^k = 2^{n-k} P(n)$$ where $P$ is a polynomial of degree $k$. For example, when $k=3$ we find $$\sum_{i=0}^3 \binom{n}{i} i^3 = 2^{n-3} (n^3 + 3n^2).$$ This means that, dividing by $n^k$, your expression of interest is $$S_n(k) \sim 2^{n-k} \left (1 \pm O \left ( \frac{1}{n} \right ) \right )$$ – HallaSurvivor May 03 '23 at 18:49
  • @HallaSurvivor Thanks for the information; I'm not quite familiar with hypergeometric functions, will check them later. As for your estimation, I think the coefficients of $P(n)=P_k(n)$ grows very fast as $k$ grows, which makes your $S_n(k)=O(1)\cdot 2^{-k}$ (when $n$ fixed) invalid (a naive lower bound is already $(1-1/n)^k$, see my original post). – Jz Pan May 05 '23 at 10:49
  • Sorry I mean that your estimation works for the case that $k$ is fixed and $n$ varies; whereas I want an estimation for $n$ fixed and $k$ varies. I'll edit my post to make it clear. – Jz Pan May 05 '23 at 11:04

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