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Is there a good upper bound for the following expression: $\sum_{k=0}^{n}\binom{n}{k} k^p$?

I see there are some questions on the closed form expression: Closed-form expression for $\sum_{k=0}^n\binom{n}kk^p$ for integers $n,\,p$ and OEIS has approximations: https://oeis.org/A072034 but is there any bounds on this quantity that are simple?

wecanfibonacciit
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  • I think this might be what you want in the question you mentioned? https://math.stackexchange.com/a/355288/823641 – Vezen BU Aug 19 '24 at 02:24
  • $$\sum_{k=0}^{n}\binom{n}{k} k^p\sim 2^{n-p}n^{p-1}\left(1+O\Big(\frac1n\Big)\right),,\text{at},,n\to\infty$$ Next asymptotic terms can also be found – Svyatoslav Aug 19 '24 at 05:11
  • How can you see this @Svyatoslav? – wecanfibonacciit Aug 20 '24 at 19:24
  • My apology for the typo - ther correct answer is $$\sum_{k=0}^{n}\binom{n}{k} k^p= 2^{n-p}n^{\color{red}{p}}\left(1+O\Big(\frac1n\Big)\right),,\text{at},,n\to\infty$$ I will post the solution – Svyatoslav Aug 23 '24 at 08:22

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We will use a heuristic approach to find the asymptotics. Let's denote

$$S(n,p)=\sum_{k=0}^{n}\binom{n}{k} k^p=n!\sum_{k=0}^{n}\frac{e^{p\ln k}}{\Gamma(k+1)\Gamma(n-k+1)},\,\,p\geqslant0$$ At $n\to\infty$ the sum has a sharp maximum near the point $k=\frac n2$. Applying the Euler-Maclaurin summation formula we note, that all terms are exponentially small compared to the first one. It means that we can switch from summation to integration. $$S(n,p)\sim n!\int_0^ne^{p\ln k-\ln\Gamma(z+1)-\ln\Gamma(n-z+1)}dz$$ Applying the Laplace' method, we find the point of maximum $z_0$ of the function $f(z)=p\ln k-\ln\Gamma(z+1)-\ln\Gamma(n-z+1)$ : $$f'(z_0)=\frac p{z_0}-\Psi(z_0+1)+\Psi(n-z_0+1)=0$$ Using the asymptotic of $\psi$-function we find $$z_0=\frac n2+\frac p2+o(1)$$ $$f''(z_0)\approx-\frac4n+\frac4{n^2}(1-p)$$ For the required level of approximation we will also need $$f^{(4)}(z_0)\approx-\frac{32}{n^3}$$ Decomposing $f(z)$ near the maximum point $z_0$ $$S(n,p)\sim\frac{n!\,z_0^p}{\Gamma(z_0+1)\Gamma(n-z_0+1)}\int_0^ne^{-\big(\frac4n-\frac4{n^2}(1-p)\big)\frac{(z-z_0)^2}2-\frac{4(z-z_0)^4}{3n^3}}dz$$ Making the substitution $z=nt$, decomposing the exponent and keeping only the terms relevant for our level of accuracy $$\sim\frac{n!\,n\,\big(\frac{n+p}2\big)^p}{\Gamma\big(\frac{n+p}2+1\big)\Gamma\big(\frac{n-p}2+1\big)}\int_{-1/2}^{1/2} e^{-2nt^2}\big(1-2(p-1)t^2\big)\left(1-\frac{4n}3t^4\right)dt$$ Making the substitution $t=\frac x{\sqrt{2n}}$ and expanding the integration to $\pm\infty$ $$S(n,p)\sim\frac{n!\,n\,\big(\frac{n+p}2\big)^p}{\Gamma\big(\frac{n+p}2+1\big)\Gamma\big(\frac{n-p}2+1\big)}\sqrt{\frac\pi{2n}}\left(1-\frac p{2n}+\frac1{4n}\right)$$ Decomposing also $$(n+p)^p\sim n^p\left(1+\frac{p^2}n\right) $$ $$\frac{n!}{\Gamma\big(\frac{n+p}2+1\big)\Gamma\big(\frac{n-p}2+1\big)}=\frac1{(n+1)B\big(\frac {n+p}2+1;\frac {n-p}2+1\big)}\sim\frac{2^{n+1}}{n+1}\sqrt{\frac n{2\pi}}\left(1-\frac{p^2-3/2}{2n}\right)$$ we get the answer: $$\boxed{\,\,S(n,p)\sim2^{n-p}n^p\left(1+\frac{p(p-1)}{2n}+O\Big(\frac1{n^2}\Big)\right)\,\,}$$

Svyatoslav
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