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Absolute convergence and uniform convergence are easy to determine for this power series. However, it is nontrivial to calculate the sum of $\large\sum \limits_{k=1}^{\infty}\frac{t^{k}}{k^{k}}$.

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According to Wolfram alpha, there is no closed form, not even for $t = 1$. –  Yuval Filmus Mar 1 '12 at 21:04
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Using Stirling's approximation, $k^k\approx e^k k!$, so it looks like this sum should go approximately like $\exp(t/e)$. –  Ben Crowell Mar 1 '12 at 21:33
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3 Answers 3

up vote 6 down vote accepted

Let's define : $\displaystyle f(t)=\sum_{k=1}^{\infty}\frac{t^{k}}{k^{k}} $

then as a sophomore's dream we have : $\displaystyle f(1)=\sum_{k=1}^{\infty}\frac 1{k^{k}}=\int_0^1 \frac{dx}{x^x}$
(see Havil's nice book 'Gamma' for a proof)
I fear that no 'closed form' are known for these series (nor integral).

Concerning an asymptotic expression for $t \to \infty$ you may (as explained by Ben Crowell) use Stirling's formula $k!\sim \sqrt{2\pi k}\ (\frac ke)^k$ to get :

$$ f(t)=\sum_{k=1}^{\infty}\frac{t^k}{k^k} \sim \sqrt{2\pi}\sum_{k=1}^{\infty}\frac{\sqrt{k}(\frac te)^k}{k!}\sim \sqrt{2\pi t}\ e^{\frac te-\frac 12}\ \ \text{as}\ t\to \infty$$

EDIT: $t$ was missing in the square root


Searching more terms (as $t\to \infty$) I got :

$$ f(t)= \sqrt{2\pi t}\ e^{\frac te-\frac 12}\left[1-\frac 1{24}\left(\frac et\right)-\frac{23}{1152}\left(\frac et\right)^ 2-O\left(\left(\frac e{t}\right)^3\right)\right]$$

But in 2001 David W. Cantrell proposed following asymptotic expansion for gamma function (see too here and the 1964 work from Lanczos) : $$\Gamma(x)=\sqrt{2\pi}\left(\frac{x-\frac 12}e\right)^{x-\frac 12}\left[1-\frac 1{24x}-\frac{23}{1152x^2}-\frac{2957}{414720x^3}-\cdots\right]$$

so that we'll compute : $$\frac{f(t)}{\Gamma\left(\frac te\right)}\sim \sqrt{t}\left(\frac {e^2}{\frac te-\frac 12}\right)^{\frac te-\frac 12}$$

and another approximation of $f(t)$ is : $$f(t)\sim \sqrt{t}{\Gamma\left(\frac te\right)}\left(\frac {e^2}{\frac te-\frac 12}\right)^{\frac te-\frac 12}$$

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Related sum. –  robjohn Sep 5 '13 at 2:58
    
Thanks for that @robjohn ! –  Raymond Manzoni Sep 5 '13 at 8:59
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I asked a similar question here. That question led Owen and me to the function you asked and we wrote up some nice (an incomplete version) properties here. To give a short answer to your question, $$\sum_{k=1}^{\infty} \frac{t^k}{k^k} = t \int_0^{1} x^{-tx}dx$$

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This is probably related to the integral

$$\int_0^1 (tx)^x dx$$

Check this and this

I don't have time to work it out now, but I'll edit in a while.


I've checked and as Sivaram points out, the integral is actually

$$t\int_0^1 x^{-tx} dx$$

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Holy cow, the ending of the episode... I mean, I knew where it's going, but still. Wow. –  Asaf Karagila Jun 2 at 4:31
    
@AsafKaragila It's 1:32 a.m. Should've gone to sleep almost 3hs ago. I am going to take up drinking. I hate George Martin. –  Pedro Tamaroff Jun 2 at 4:33
    
If you'd read all the spoilers/the books, you wouldn't be surprised. You'd be amazed at the graphical description of the fight. –  Asaf Karagila Jun 2 at 4:43
    
@AsafKaragila I actually read the complete scene from the book when I finished, though I never read the book! –  Pedro Tamaroff Jun 2 at 4:47
    
Hah, yeah, me too. I liked the TV adaptation better. –  Asaf Karagila Jun 2 at 4:49
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