Asymptotics of the sum $\sum_{n=1}^\infty \frac{x^n}{n^n}$

Question

How does the sum $$f(x)=\sum_{n=1}^\infty \frac{x^n}{n^n}$$ behave asymptotically as $x\to\infty$? It appears that $f(x)$ asymptotically dominates any polynomial and is dominated by any exponential, so we might consider $\log f(x)$ rather than $f(x)$.

I apologize for having no work to show on this problem; I have no idea how to begin tackling a problem regarding the asymptotics of a function given its power series (of which there is no hope of evaluating in closed form). Hopefully an answer will provide me with some tools for doing so.

Answer 1

Written from an airport. This is somewhat sketchy when comparing solutions to differential equations, but hopefully not too much for you to fill the gaps.

The main idea: bounding $f$ via differential partial equation.

We have $$ f'(x) = \sum_{n=1}^\infty \frac{x^{n-1}}{n^{n-1}} = \sum_{n=0}^\infty \frac{x^{n}}{(n+1)^{n}} = 1+\sum_{n=1}^\infty \frac{\frac{x^{n}}{n^n} }{\left(1+\frac{1}{n}\right)^{n}} > 1+\frac{1}{e}\sum_{n=1}^\infty \frac{x^{n}}{n^n} = 1+\frac{1}{e}f(x) \tag{1} $$ so in particular $$ f' > 1+\frac{1}{e}f \tag{2} $$ Since $f(0) = 0$, and the solution to $g' = 1+e^{-1}g$ with $g(0)=0$ is given by $g(x) = e^{x/e+1}-e$, we have $$ f(x) \geq e^{x/e+1}-e > 2e^{x/e} , \qquad x>4\tag{3} $$ ($x>4$ for the second inequality to kick in). Now, from $(1)$ we also have $$ f' < 1+f \tag{4} $$ (we can even improve this to $f' < 1+\frac{2}{3}f$), which this time gives $$ f(x) \leq e^x - 1\tag{5} $$

Overall, for $x>4$, $$ 2e^{x/e} \leq f(x) \leq e^x - 1 \tag{6} $$ which provides a loose estimate of the asymptotic growth of $f$: namely, $\boxed{f(x) = e^{\Theta(x)}}$.

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Further: Improving (slightly) on the lower bound on $\log f$ by the low-order terms, and improving on the constant in the main asymptotics of the upper bound of $\log f$.

I will show $$ h(x) \leq f(x) \leq g(x) \tag{7} $$ where $$ \begin{align} \log h(x) &= \frac{1}{e}x + 4 - \log\frac{32}{3} + o(1) \tag{8}\\ \log g(x) &= \frac{256}{625}x + O(1) \tag{9} \end{align} $$ (note that $\frac{256}{625} \approx \frac{1}{e}+0.04$). Moreover, this can be improved by the same method, pushing to more accurate estimates, but this will get uglier. (One can also push the Taylor expansion above further, based on (12) and (13). I stopped at $o(1)$).

The observation is that for the upper and lower bound, we bounded uniformly the coefficients by $$ \forall n \geq 1\, \qquad \frac{1}{n^n} \leq \frac{1}{\left(1+\frac{1}{n}\right)^{n}} \cdot \frac{1}{n^n} \leq \frac{1}{e}\cdot \frac{1}{n^n} $$ to obtain the two corresponding differential equations. We can do better by handling the first few terms more tightly. Namely, we have $$ \left(1+\frac{1}{n}\right)^n = \begin{cases} \frac{1}{2} & n=1\\ \frac{4}{9} & n=2\\ \frac{27}{64} & n=3\\ \frac{256}{625} & n=4 \end{cases} $$ (and, of course, $\left(1+\frac{1}{n}\right)^n$ is decreasing to $1/e$). Thus, we can leverage this and solve instead the following two differential equations for $h$ and $g$: \begin{align} h'(x) &= 1 + \left(\frac{1}{2} - \frac{1}{e}\right) x + \left(\frac{4}{9} - \frac{1}{e}\right) \frac{x^2}{4} + \left( \frac{27}{64} - \frac{1}{e}\right) \frac{x^3}{27} + \frac{1}{e}h(x)\tag{10}\\ g'(x) &= 1 + \left(\frac{1}{2} - \frac{256}{625}\right) x + \left(\frac{4}{9} - \frac{256}{625}\right) \frac{x^2}{4} + \left( \frac{27}{64} - \frac{256}{625}\right) \frac{x^3}{27} + \frac{256}{625}g(x)\tag{11} \end{align} subject to $h(0)=g(0)=0$. Solving those gives a nasty expression, \begin{align} h(x) &= \frac{3}{32} e^{4 + \frac{1}{e}x} + \left(\frac{1}{27} - \frac{e}{64}\right) x^3 + \left(\frac{1}{4} - \frac{3 e^2}{64}\right) x^2 + \left(1 - \frac{3e^3}{32}\right) x -\frac{3e^4}{32} \tag{12} \\ g(x) &= \frac{457763671875}{137438953472}e^{\frac{256}{625}x} - \frac{491}{442368}x^3 - \frac{123299}{4194304}x^2 - \frac{195550963}{536870912}x -\frac{457763671875}{137438953472} \tag{13} \\ \end{align} leading to the claimed (8) and (9).

Below, a plot illustrating those approximations:

Answer 2

There is an analogue of Laplace's method which works for sums. $n \ln(x/n)$ attains the maximum at $n = x/e$. Writing the exponent as $n \ln(x/n) = x/e - x \xi^2$, computing the expansion of $n'(\xi)$ at $\xi = 0$ and extending the integration range to $(-\infty, \infty)$, we obtain $$\frac {n'(\xi)} x = \sqrt{\frac 2 e} + c_1 \xi - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + c_3 \xi^3 + O(\xi^4), \quad \xi \to 0,\\ \sum_{n \geq 1} \frac {x^n} {n^n} = \int_{-\infty}^\infty x \left( \sqrt{\frac 2 e} - \frac 1 6 \sqrt{\frac e 2} \,\xi^2 + O(\xi^4) \right) e^{x/e - x \xi^2} d\xi = \\ \sqrt{\frac \pi 2} \,e^{x/e} \left( 2\sqrt{\frac x e} - \frac 1 {12} \sqrt{\frac e x} + O(x^{-3/2}) \right), \quad x \to \infty,$$ which gives $\ln f(x)$ with an error of order $O(x^{-2})$.

Answer 3

Similar to Maxim:

$$ \begin{align} f(a)&=\sum_{n=1}^\infty \frac{a^n}{n^n}=\sum_{n=1}^\infty e^{n \log(a/n)}\\ &\approx \int_1^\infty e^{t \log(a/t)} dt \\ &= a \int_0^a \frac{1}{u^2} e^{a \log(u) /u} du\\ &= a \int_0^a h(u) e^{a g(u)} du\\ &\approx a \sqrt{\frac{2 \pi}{a |g''(u_0)|} } h(u_0) e^{a g(u_0)} \end{align} $$

where we've used Laplace's approximation (assuming $a \gg e$) to $h(u) =\frac{1}{u^2}$ and $g(u)=\log(u)/u$, with $u_0=e$ , $g''(u_0)=-1/e^3$ . Then the approximation gives

$$f(a)\approx \sqrt{2 \pi a} \exp( a/e-1/2)$$

or

$$\log f(a)\approx \frac{a}{e} + \frac{1}{2}\log(a) + \frac{1}{2}(\log(2 \pi)-1) $$

I've not done the strict asyptotical analysis, but it looks as if the error is $o(1)$. Some numerical values

 a      log(f(a))    aprox      abs error 
3.0     1.896554    2.071883    0.175329
5.0     2.984687    3.063055    0.078368
7.5     4.150135    4.185486    0.035350
10.0    5.229637    5.249025    0.019389
20.0    9.268130    9.274393    0.006264
30.0    13.151944   13.155920   0.003976
50.0    20.766590   20.768922   0.002332
75.0    30.167102   30.168641   0.001539
100.0   39.508319   39.509468   0.001149
200.0   76.643415   76.643985   0.000570
300.0   113.634283  113.634662  0.000379
500.0   187.465736  187.465963  0.000227
750.0   279.638405  279.638556  0.000151
1000.0  371.752144  371.752257  0.000113

Answer 4

If we apply the identity $$ \frac{1}{{n^n }} = \frac{1}{{(n - 1)!}}\int_0^{ + \infty } {t^{n - 1} \mathrm{e}^{ - nt} \mathrm{d}t} , \quad n\geq 1, $$ we obtain the integral representation $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} = x\int_0^{ + \infty } {\mathrm{e}^{ - xp(t)} q(t) \mathrm{d}t} , $$ with $p(t) = - t\mathrm{e}^{ - t}$ and $q(t)=\mathrm{e}^{ - t}$. The $p(t)$ has a sole, simple saddle point on the path of integration at $t=1$. Employing the saddle point method, we find $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n^n }}} \sim \mathrm{e}^{z} \sqrt {2\pi z} \sum\limits_{k = 0}^\infty {\frac{{a_k }}{{z^k }}} = \mathrm{e}^{z} \sqrt {2\pi z} \left( {1 - \frac{1}{{24z}} - \frac{{23}}{{1152z^2 }} - \frac{11237}{414720 z^3}-\ldots } \right) $$ as $x\to +\infty$, with $z=x/\mathrm{e}$ and $$ a_k = \frac{1}{{2^k k!}}\left[ {\frac{{\mathrm{d}^{2k} }}{{\mathrm{d}t^{2k} }}\left( {\mathrm{e}^{ - t} \left( {\frac{1}{2}\frac{{t^2 }}{{1 - (t + 1)\mathrm{e}^{ - t} }}} \right)^{k + 1/2} } \right)} \right]_{t = 0} . $$ An alternative expression for the coefficients involving the Bernoulli polynomials $B_j(x)$ is $$ a_k = \frac{1}{k!}\left[ \frac{\mathrm{d}^k }{\mathrm{d}t^k}\exp \left( \sum\limits_{j = 1}^k B_{j + 1}\! \left( \tfrac{3}{2} - k \right)\frac{t^j}{j(j + 1)} \right) \right]_{t = 0} . $$ I omit the proof.

Answer 5

Here is a simple argument that gives simple bounds. The idea is just to apply Stirling's approximation in reverse. Namely, using the crude Stirling inequalities

$$e \left( \frac{n}{e} \right)^n \le n! \le en \left( \frac{n}{e} \right)^n$$

we get the inequalities

$$e^{n-1} (n-1)! \le n^n \le e^{n-1} n!$$

which give, for $x \ge 0$, the upper and lower bounds

$$\boxed{ e^{\frac{x}{e} + 1} - e = \sum_{n \ge 1} \frac{x^n}{e^{n-1} n!} \le f(x) \le \sum_{n \ge 1} \frac{x^n}{e^{n-1} (n-1)!} = x e^{\frac{x}{e}} }.$$

This is already enough to establish the asymptotic behavior of $f(x)$ as $x \to \infty$ up to a factor of $x$, which in particular gives $\boxed{ \ln f(x) \sim \frac{x}{e} + O(\ln x) }$.