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

  • $\begingroup$ According to Wolfram alpha, there is no closed form, not even for $t = 1$. $\endgroup$ – Yuval Filmus Mar 1 '12 at 21:04
  • 2
    $\begingroup$ Using Stirling's approximation, $k^k\approx e^k k!$, so it looks like this sum should go approximately like $\exp(t/e)$. $\endgroup$ – Ben Crowell Mar 1 '12 at 21:33

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


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


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

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

  • $\begingroup$ Holy cow, the ending of the episode... I mean, I knew where it's going, but still. Wow. $\endgroup$ – Asaf Karagila Jun 2 '14 at 4:31
  • $\begingroup$ @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. $\endgroup$ – Pedro Tamaroff Jun 2 '14 at 4:33
  • $\begingroup$ 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. $\endgroup$ – Asaf Karagila Jun 2 '14 at 4:43
  • $\begingroup$ @AsafKaragila I actually read the complete scene from the book when I finished, though I never read the book! $\endgroup$ – Pedro Tamaroff Jun 2 '14 at 4:47
  • $\begingroup$ Hah, yeah, me too. I liked the TV adaptation better. $\endgroup$ – Asaf Karagila Jun 2 '14 at 4:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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