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I'm no analyst, so when a student in the class to whom I was teaching some elementary theory of (power) series, asked about this:

$\displaystyle{\sum_{n=0}^\infty\frac{x^n}{n^n}=1+x+\left(\frac{x}{2}\right)^2+\left(\frac{x}{3}\right)^3+\left(\frac{x}{4}\right)^4+\cdots}$

(assuming $0^0=1$), I had no idea. It didn't look like anything I recognized, and a play about with some derivatives gave me no useful information.

Is this a known function, and does this series admit of an explicit formula?

Alasdair
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  • You can calculate a good value at $x=1,$ say, and put the decimal digits at http://oeis.org/search?q=7%2C1%2C8%2C2%2C8%2C1%2C8%2C2%2C8&language=english&go=Search or put as a real number at http://isc.carma.newcastle.edu.au/standard or http://isc.carma.newcastle.edu.au/advanced – Will Jagy Oct 22 '13 at 01:59
  • Yep, I already tried that (putting the digits of x=1 into OEIS), and got returned that it was the value of the sum of 1/n^n from 0 to infinity. The inverse symbolic calculator at newcastle gives the same. – Alasdair Oct 22 '13 at 02:16
  • That suggests that this function has no name of its own....Also, you don't come across entire functions that often. Modular functions, what have you, typically meromorphic or worse; just not that common to study something with infinite radius of convergence other than $e^z$ – Will Jagy Oct 22 '13 at 02:18
  • Talking about useful information, did you determine the Interval of Convergence? That is doable with the root test, I believe – imranfat Oct 22 '13 at 02:22
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    @imranfat Each coefficient is much, much smaller than the corresponding coefficient of $e^x$. Thus this converges everywhere, very quickly. – davidlowryduda Oct 22 '13 at 02:25
  • probably something to be learned from the reference books for http://en.wikipedia.org/wiki/Entire_function although the only similar series displayed is not given a name. – Will Jagy Oct 22 '13 at 02:52
  • This power series does have a real root at $x\approx-1.403761051217752$. Needless to say, all real roots are necessarily negative...but I think this is the only real root. – Alex Nelson Dec 29 '13 at 04:59

2 Answers2

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To start analyzing this series, you could consider using Sterling's approximation $n!\approx \left(\frac ne\right)^n\sqrt {2\pi n}$:

$$\sum_{n=0}^\infty\left(\frac xn\right)^n\approx \sqrt{2\pi}\sum_{n=0}^\infty \frac {\sqrt n({x\over e})^n}{n!}\ge \sqrt{2\pi}\sum_{n=0}^\infty \frac {({x\over e})^n}{n!}=e^{\frac xe} \sqrt{2\pi}$$

Of course this is only a lower limit to the approximated value, and that $\sqrt n$ term still needs to be dealt with...

abiessu
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I had to do a lot of research on this, which was my benefit :)

Outline. Basically, I will use the Sophomore's Dream to deduce an integral expression for the answer.

Lemma. We see $$\int^{1}_{0}x^{n}\ln(x)^{n}\,\mathrm{d}x = (-1)^{n}(n+1)^{-(n+1)}n!\tag{1}$$

Proof. We change variables writing $$\tag{2a}x=\exp\bigl(-u/(n+1)\bigr)$$ which lets us rewrite (1) as $$\int^{1}_{0}x^{n}\ln(x)^{n}\,\mathrm{d}x = (-1)^{n}(n+1)^{-(n+1)}\int^{\infty}_{0}u^{n}e^{-u}\,\mathrm{d}u\tag{2b}$$ Observe the integral on the right hand side is precisely $n!$ (thanks to the Gamma function). And that concludes the proof for our lemma. QED.

Theorem. We claim $$f(t) = 1+\sum^{\infty}_{n=1}\left(\frac{t}{n}\right)^{n} = 1 + t\int^{1}_{0}x^{-xt}\,\mathrm{d}x.\tag{3}$$

Proof. We end up rewriting the integrand on the right hand side $$x^{-xt} = e^{-xt\ln(x)} = \sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}x^{n}\ln(x)^{n}.\tag{4a}$$ We plug this back into the integral $$\int^{1}_{0}x^{-xt}\,\mathrm{d}x= \int^{1}_{0}\sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}x^{n}\ln(x)^{n}\,\mathrm{d}x.\tag{4b}$$ Swap the sum and integral $$\int^{1}_{0}\sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}x^{n}\ln(x)^{n}\,\mathrm{d}x =\sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}\int^{1}_{0}x^{n}\ln(x)^{n}\,\mathrm{d}x.\tag{4c}$$ We use our lemma to rewrite the right hand side as \begin{align} \sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}\int^{1}_{0}x^{n}\ln(x)^{n}\,\mathrm{d}x &= \sum^{\infty}_{n=0}\frac{(-t)^{n}}{n!}\left((-1)^{n}(n+1)^{-n+1}n!\right)\\ &= \sum^{\infty}_{n=0}\frac{t^{n}}{(n+1)^{n+1}}.\tag{4c} \end{align} Then we just fiddle with arithmetic (multiply by $t$, and add 1) to get the series in question. That concludes our proof for the theorem. QED.

Remark. There is no closed form expression for the integral that I am aware of. Perhaps either the OP or some other user knows some fancy-pants way to evaluate the integral, but I do not know of one readily available :(

Alex Nelson
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