# “Closed” form for $\sum \frac{1}{n^n}$

Earlier today, I was talking with my friend about some "cool" infinite series and the value they converge to like the Basel problem, Madhava-Leibniz formula for $\pi/4, \log 2$ and similar alternating series etc.

One series that popped into our discussion was $\sum\limits_{n=1}^{\infty} \frac{1}{n^n}$.

Proving the convergence of this series is trivial but finding the value to which converges has defied me so far. Mathematica says this series converges to $\approx 1.29129$.

I tried Googling about this series and found very little information about this series (which is actually surprising since the series looks cool enough to arise in some context).

We were joking that it should have something to do with $\pi,e,\phi,\gamma$ or at the least it must be a transcendental number :-).

My questions are:

1. What does this series converge to?
2. Does this series arise in any context and are there interesting trivia to be known about this series?

I am actually slightly puzzled that I have not been able to find much about this series on the Internet. (At least my Google search did not yield any interesting results).

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Did you try Plouffe's Invertor? –  Aryabhata Feb 10 '11 at 7:49
@Moron: Nice website. Through the website, I was only able to get the numerical value up to 1000 digits. pi.lacim.uqam.ca/piDATA/sumnn.txt –  user17762 Feb 10 '11 at 7:58
I asked a related question on MO: mathoverflow.net/questions/33397/… . It is possible to make "deformed" trigonometric functions by replacing $e^{z}$ with $M(z)$, but so far I haven't found references to them in lists of special functions. –  deoxygerbe Feb 10 '11 at 9:32
@deoxygerbe: Very Interesting! I was thinking of a function on similar lines and study its behavior. The function actually looks a cool function. It converges for all $z$ and is holomorphic. An interesting function to play around with! –  user17762 Feb 10 '11 at 9:40
@Sivaram just emailed you what I wrote up about them before. –  deoxygerbe Feb 10 '11 at 9:47

## 4 Answers

Certainly you need to check out Sophomore's Dream.

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Nice result! –  user17762 Feb 10 '11 at 8:01
I wonder thy they call it like that. Very few sophomores might ever come up with such an identity... –  Mariano Suárez-Alvarez Feb 10 '11 at 9:09
Right. It's just a clever saying, especially considering that the freshman's dream happens all the time. –  Mitch Feb 10 '11 at 15:09

This series turns out to be equal to the integral of $1/x^x$ from $0$ to $1$. Sorry I am bad with the notation needed for math symbols here; but if you rewrite $x^{-x}$ as the exponential of $x\log x$, you can expand it as a taylor series and integrate term by term; to find it equals your sum. This doesn't give an explicit value; but I think it's a pretty cool identity.

EDIT: Just putting what you said into symbols, probably for the fun of it.

$$\int_0^1 x^{-x} dx=\int_0^1 e^{-x \log x} dx$$

$$e^{-x \log x} =\sum_{k=0}^\infty (-1)^k \frac{x^k\log^kx}{k!}$$

But since

$$\int_0^1 x^k \log^k x dx =(-1)^k \frac{k!}{(k+1)^{k+1}}$$

we get

$$\int_0^1 x^{-x} dx= \sum_{k=0}^\infty \frac{1}{(k+1)^{k+1}}=\sum_{k=1}^\infty \frac{1}{k^k}$$

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If you want superscripts or subscripts of more than one character, you need to enclose them in {}. –  Ross Millikan Feb 10 '11 at 15:01
1969 Putnam exam, problem A4: Show that $\int_0^1x^x\,dx=\sum_{n=1}^{\infty}(-1)^{n+1}n^{-n}$. –  Gerry Myerson May 11 '11 at 1:55

Liouville's theorem implies that if your constant is irrational then it is in fact transcendental. Unfortunately, the trivial way of proving irrationality doesn't work. On second thought, the same problem also prevents us from applying Liouville's theorem!

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I am unable to immediately see that if the constant is irrational then it has to be a Liouville number. Could you throw light on the proof? –  user17762 Feb 11 '11 at 2:57
@Sivaram: You have a point... –  Yuval Filmus Feb 11 '11 at 4:45
To apply Liouville's theorem (or the much stronger Roth version) you need an infinite sequence of rational approximations. The smallest upper bound I get on the denominators of the obvious sequence of rational approximations ($\sum_{n=1}^l \frac{1}{n^n}$) is $q = \prod_{i=1}^n i^i$, whereas the best bound I get for $\left| \frac{p}{q} - \alpha \right|$ is $\frac{1}{n^n} \sum_{k=1}^\infty \frac{1}{k^k}$. That's huge compared to $1/q^r$ so the approximation theory doesn't apply at all - unless you produce a much better sequence of rationals and bound them very tightly. –  quanta May 7 '11 at 21:38

This has been asked here as well:

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