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The function $F(x) = \sum_{0}^{\infty} x^n/n^n$ may be familiar to many readers as an example sometimes used when teaching tests for absolute convergence of entire functions defined by power series. I know of no name for it, nor any use for it aside from pedagogical, so this is a pure curiosity question which I hope is acceptable. The function seems to have one real zero around x = -1.40376; a single extremum, a minimum around x = -5.71837; and then to approach the x-axis from below asymptotically as x goes to negative infinity. Is this true?

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Because the series is absolutely convergent, it follows that $F(x)$ is a holomorphic function. If $x$ is real and $x>0$, then it is clear that $F(x)$ and $F'(x) = \sum_{n \geq 1} (x/n)^{n-1}$ are both positive. And if $x$ is real and $x \leq 0$, we can use the alternating series bounds to estimate $F(x)$. Then using a computer, it seems that your claims are correct. –  Pat Devlin Oct 8 '12 at 19:24
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Perhaps we can use Stirling's formula to get bounds on $1/n^n$ in terms of $n!$. Then for all $x$, we could perhaps see that $F(x)$ is bounded by something on the order of $e^{x}$. This would prove that $\lim_{x \to - \infty} F(x) = 0$. –  Pat Devlin Oct 8 '12 at 21:08
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Notice Pat Devlin's expression suggests that F'(x) is close to eF(x). Gerhard "Perhaps Taking Glasses Off Helps" Paseman, 2012.10.08 –  Gerhard Paseman Oct 8 '12 at 21:43
    
@Pat: Actually, even a small perturbation of a single coefficient of the power series produces a catastrophic effect on the limit of $F(x)$ for $x\to-\infty$. So one may ask what properties (quite rigid, in any case) on these coefficients make $F(x)\to0$ for $x\to-\infty$, since both $1/n^n$ and $1/n!$ works. –  Pietro Majer Oct 10 '12 at 9:27
    
Your observation converts this into a research problem--to characterize Majer sequences, where a Majer sequence is a sequence of positive terms such that $\sum s_n x^n$ is entire and is asymptotic to the x-axis as $x \rightarrow -\infty$. As you note, not any superficially similar sequence will do, by any means--a good example to consider is BesselI(0, 2 sqrt(x)), which has $s_n = 1/(n!)^2$. It seems likely we would need to start from the functions themselves instead of trying to characterize the property in terms of sequences, but one wonders. –  Gene Ward Smith Oct 10 '12 at 14:46
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2 Answers 2

up vote 23 down vote accepted

[Edited to outline the end of the argument that $f(-M) \rightarrow 0$ (and to correct a few typos etc. while I'm at it)]

Yes, $F(x) \rightarrow 0$ from below as $x \rightarrow -\infty$. The convergence is slow, and precise asymptotic analysis seems to be somewhat annoying because it involves the lower branch of the Lambert W function.

The massive cancellations in $\sum_{n=0}^\infty x^n/n^n$ for $x \rightarrow -\infty$ can be tamed by the familiar device of writing $$ \frac1{n^n} = \frac1{(n-1)!} \int_0^\infty t^{n-1} e^{-nt} dt $$ for $n=1,2,3,\ldots$. Multiplying by $x^n$, summing over $n>0$, and restoring the $n=0$ term $x^0/$"$0^0$"$=1$ yields $$ f(x) = 1 + x \int_0^\infty e^{txe^{-t}} e^{-t} dt. $$ Hence if $x=-M$ then $$ f(x) = f(-M) = 1 - M \int_0^\infty e^{-Mte^{-t}} e^{-t} dt, $$ and as $M \rightarrow +\infty$ the integral naturally splits into the parts $t \leq 1$ where $t e^{-t}$ is increasing and $t \geq 1$ where $t e^{-t}$ is decreasing. We let $u = t e^{-t}$, so the integrand becomes $e^{-Mu} du/(1-t)$. For $t<1$ we use Abel's power series $t = \sum_{m=1}^\infty m^{m-1} u^m/m!$ to expand the integral in an asymptotic series: $$ \int_0^1 e^{-Mte^{-t}} e^{-t} dt \sim \frac1M + \frac1{M^2} + \frac{2^2}{M^3} + \frac{3^3}{M^4} + \frac{4^4}{M^5} + \cdots $$ which is already enough to get $f(-M) < 0$ for large $M$. [Curiously the asymptotics of $\sum_{n=0}^\infty (-M)^n/n^n$ have led us to the divergent series $\sum_{n=1}^\infty n^n/M^n$.]

But the resulting bound $f(-M) < -1/M$ underestimates $|f(-M)|$: numerically $f(-100) \simeq -.1826$, $\phantom.$ $f(-1000) \simeq -.1180$, and $\phantom.$ $f(-10000) \simeq -.0899$, suggesting that $f(-M)$ decays only as $-1/\log M$ or so. The reason must be the $t>1$ part of the integral. On this part, $t = \log(1/u) + \log\log(1/u) + o(1)$ as $t \rightarrow \infty$, so the integral behaves to first order like $\int_0^{1/e} e^{-Mu} du / \log(1/u)$. Now $\log(1/u) \rightarrow 0$ as $u \rightarrow 0+$, but the convergence is slower than any positive power of $u$. Therefore, the integral is $o(1/M)$, which completes the argument that $f(x) \rightarrow 0$ as $x \rightarrow -\infty$; but the integral is not $O(1/M^\theta)$ for any $\theta > 1$, so $f(-M)$ decays slower than any positive power of $M$.

A more thorough asymptotic analysis of the $t>1$ integral as $M \rightarrow \infty$ looks routine but unpleasant, so I'll stop at this point; perhaps somebody else here will be interested in pursuing it further.

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It's a really nice presentation. BTW,i believe you mean $e^{-nt}$ in the first integral above. –  user23078 Oct 9 '12 at 1:11
    
Thanks! Yes, $e^{-nt}$, not $e^{nt}$; also, restoring the $n=0$ (not "$x=0$") term. Will fix this and make a few other edits tomorrow. –  Noam D. Elkies Oct 9 '12 at 2:09
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I wish to add a remark to Noam D. Elkies' beautiful answer. From the integral representation for $f$, putting $e^{-t}=s$ in the integral, $$f(-x)=1-x\int_0^\infty e^{-xte^{-t}} e^{-t}dt = 1-x\int_0^1 s^{sx}ds\, ,$$ so that, for $x\to \infty$, $ f(-x)=o(1)$ is equivalent to $$\int_0^1 xu(s)^xds=1+o(1)\, ,$$ where $u\in C([0,1])$ is the function $u(s):=s^s$. As a matter of fact, since $0\le u(s)\le 1$ for all $s$ and $u(s)=1$ only for $s=0$ or $s=1$, it turns out that the limit only depends on $u'(0)$ and $u'(1)$.

Since $u'(1)=1$, for any $\lambda < 1 < \mu$ there exists a $b < 1$ such that for all $s\in [b,1]$ there holds $$1+\mu(s-1) \le u(s)\le 1+\lambda(s-1)\, ,$$ so that $$x\big(1+\mu(s-1)\big)^x \le xu(s)^x\le x\big(1+\lambda(s-1)\big)^x\, .$$ Similarly, since $u'(0)=-\infty$, for any $\nu > 0$ there exists a $a > 0$ such that for all $s\in [0,a]$ $$u(s)\le1-\nu s\, ,$$ so $$xu(s)^x\le x\big( 1-\nu s\big) ^ x \, .$$

Moreover, since on any interval $[a,b]\subset\subset(0,1)$ the function $u$ is bounded away from $1$, it is clear that $\int_a^b xu(s)^xds=o(1)$ by uniform convergence to $0$.

Integrating over $s\in [ 0,1]$, and recalling that $\lambda < 1 < \mu$ and $\nu > 0$ were arbitrary, the inequalities above plainly give

$$\int_0^1 xu(s)^xds=\int_0^a x u(s)^xds+\int_a^b xu(s)^xds+\int_b^1 xu(s)^xds=1+o(1) \, ,$$ for $x\to \infty$.

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This is a nice expression for $f(x)$; in retrospect the integral formula I gave is just a direct generalization of the familiar curiosity about $\int_0^1 s^{\pm s} ds$. Still it seems that integration with respect to $t = -\log s$ is more convenient for the asymptotic analysis. –  Noam D. Elkies Oct 10 '12 at 3:50
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