Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
According to Wolfram alpha, there is no closed form, not even for $t = 1$. – Yuval Filmus Mar 1 '12 at 21:04
2  
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

3 Answers

up vote 5 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 : $$f(t)\sim \sqrt{t}{\Gamma\left(\frac te\right)}\left(\frac {e^2}{\frac te-\frac 12}\right)^{\frac te-\frac 12}$$

share|improve this answer

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, $$\sum_{k=1}^{\infty} \frac{t^k}{k^k} = t \int_0^{1} x^{-tx}dx$$

share|improve this answer

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.