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Consider the OEIS sequence A031971, which is defined as: $$a_n=\sum\limits_{k=1}^n k^n\quad\color{gray}{(1,\,5,\,36,\,354,\,4425,\,67171,\,1200304,\,.\!.\!.\!)}\tag{1}$$ I'm interested in the asymptotic behavior of $a_n$ for $n\to\infty$.

Empirically, it appears that $$a_n\stackrel{\color{gray}?}\sim\frac{e}{e-1}\,n^n\cdot\left(1-\frac{e+1}{2\,(e-1)^2}\,n^{-1}+c\,n^{-2}+\mathcal O\!\left(n^{-3}\right)\right),\tag{2}$$ where $c\approx0.6310116...$ (I haven't found a plausible closed form it). The leading term $\frac{e}{e-1}\,n^n$ is given in the OEIS.

How can we prove this formula and find higher terms in it?

Vladimir Reshetnikov
  • 47,122

2 Answers2

11

Let us write $$ \sum\limits_{k = 1}^n {k^n } = n^n \sum\limits_{k = 0}^{n - 1} {\left( {1 - \frac{k}{n}} \right)^n } . $$ Now by power series expansions \begin{align*} \left( {1 - \frac{k}{n}} \right)^n & = \exp \left( {n\log \left( {1 - \frac{k}{n}} \right)} \right) = \exp \left( { - n\sum\limits_{j = 1}^\infty {\frac{{k^j }}{j}\frac{1}{{n^j }}} } \right) \\ & = \mathrm{e}^{ - k} - \mathrm{e}^{ - k} \frac{{k^2 }}{2}\frac{1}{n} + \mathrm{e}^{ - k} \left( {\frac{{k^4 }}{8} - \frac{{k^3 }}{3}} \right)\frac{1}{{n^2 }} - \ldots \,. \end{align*} Substituting back to the original sum, extending that sum to $n=\infty$, and using closed forms of $\sum_{k=0}^\infty k^px^k$ (with $x=\mathrm{e}^{-1}$), we deduce \begin{align*} \sum\limits_{k = 1}^n {k^n } & \sim n^n \left( {\frac{\mathrm{e}}{{\mathrm{e} - 1}} - \frac{{\mathrm{e}^2 + \mathrm{e}}}{{2(\mathrm{e} - 1)^3 }}\frac{1}{n} - \frac{{\mathrm{e}(5\mathrm{e}^3 - 9\mathrm{e}^2 - 57\mathrm{e} - 11)}}{{24(\mathrm{e} - 1)^5 }}\frac{1}{{n^2 }} + \ldots } \right) \\ & =\frac{\mathrm{e}}{{\mathrm{e} - 1}}n^n \left( {1 - \frac{{\mathrm{e} + 1}}{{2(\mathrm{e} - 1)^2 }}\frac{1}{n} - \frac{{5\mathrm{e}^3 - 9\mathrm{e}^2 - 57\mathrm{e} - 11}}{{24(\mathrm{e} - 1)^4 }}\frac{1}{{n^2 }} + \ldots } \right) \end{align*} as $n\to +\infty$. You can obtain more terms of the expansion if you like, the process should be clear.

Gary
  • 31,845
2

The leading behavior may be understood by rendering

$\Sigma_{k=0}^n k^n = n^n\Sigma_{k=0}^n(k/n)^n = n^n\Sigma_{m=0}^n(1-m/n)^n.$

Here $m=n-k$. Then the terms in the last sum successively approach $e^{-m}$ as $n$ increases without bound, rendering the asymptotic result

$\Sigma_{k=0}^n k^n\text{ ~ } n^n\Sigma_{m=0}^\infty e^{-m} = (n^n)[e/(e-1)].$

Oscar Lanzi
  • 39,403