Absolute convergence and uniform convergence are easy to determine for this power series. However, it is nontrivial to calculate the sum of $\large\sum \limits_{k=1}^{\infty}\frac{t^{k}}{k^{k}}$.
What's the sum of $\sum \limits_{k=1}^{\infty}\frac{t^{k}}{k^{k}}$?
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2Using Stirling's approximation, $k^k\approx e^k k!$, so it looks like this sum should go approximately like $\exp(t/e)$. – 2012-03-01
3 Answers
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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0[Rel$a$ted sum](http://math.stackexchange.com/a/205896). – 2013-09-05
This is probably related to the integral
$\int_0^1 (tx)^x dx$
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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0@AsafKaragila The quality of the show has been increasing, methinks. Still not over the Viper being gone =(. Loved his character. – 2014-06-02