## To what does this series evaluate?

For the discussion of math. Duh.

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Mike_Bson
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### To what does this series evaluate?

This is just a problem I have been curious about:

$\sum_{k=1}^{\infty} k^{-k}$

I've been trying to evaluate this for the exact value for a few hours. I know this converges to about 1.291285997, but I'm trying to find the exact value; does anyone know the answer, or a way to find the answer? I've tried finding partial sums, but the results have come out unhelpful. Any help would be appreciated, thanks.

antonfire
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### Re: To what does this series evaluate?

Unfortunately, "most" sums don't have closed form expressions. It's likely (though probably hasn't been shown) that this one doesn't. Chances are, the best you can do is relate it to other esoteric expressions.
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mdyrud
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### Re: To what does this series evaluate?

Whenever I run into something like this that I can't figure out, my first step is to check WolframAlpha. For this one, it agrees that there is no closed form expression. It's not a good method to follow if you are working on homework problems giving practice for an upcoming test, but if you're just curious, check it out.

pizzazz
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### Re: To what does this series evaluate?

Isn't the very first identity on the wiki page the sum the OP is asking about?

jestingrabbit
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### Re: To what does this series evaluate?

pizzazz wrote:Isn't the very first identity on the wiki page the sum the OP is asking about?

Yes, but it doesn't provide a simple expression for the value, like pi^2/6, which is probably what the OP was asking about.
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Mike_Bson
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### Re: To what does this series evaluate?

antonfire wrote:Unfortunately, "most" sums don't have closed form expressions. It's likely (though probably hasn't been shown) that this one doesn't. Chances are, the best you can do is relate it to other esoteric expressions.

Hm, the relationship with that definite integral is indeed interesting. Couldn't you use the fundamental theorem to express the number in a closed form? I know Wolfram can't find the antiderivative of x^-x, but does it exist?

There's probably something big I'm missing. . . .

Qaanol
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### Re: To what does this series evaluate?

Mike_Bson wrote:Hm, the relationship with that definite integral is indeed interesting. Couldn't you use the fundamental theorem to express the number in a closed form? I know Wolfram can't find the antiderivative of x^-x, but does it exist?

Yes, it exists, along the positive reals anyway. Just define [imath]g(0) = 0[/imath] and for [imath]x > 0[/imath] set
$g(x) = \lim_{a\rightarrow 0^+}\int_a^x g(t)dt.$

For our current purposes, the function [imath]f(x) = x^{-x}[/imath] can be written [imath]f(x) = e^{-x\log(x)}[/imath]. Taking the principle branch of the natural logarithm, that function is analytic away from the non-negative real line. In particular, it has a local power series representation. Taking that power series about [imath]x = 1[/imath] will give a radius of convergence of 1. Integrating the power series termwise yields an antiderivative F on that domain of convergence. In particular, this domain includes the open real interval (0, 2). Taking the constant of integration as 0, we get [imath]F(1) = 0[/imath]. We may then define
$F(0) = \lim_{x\rightarrow 0^+}F(x)$
where the limit is taken along the positive reals. Indeed we may center our power series at any positive real number p, and get an interval of convergence (0, 2p).
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### Re: To what does this series evaluate?

Mike_Bson wrote:Hm, the relationship with that definite integral is indeed interesting. Couldn't you use the fundamental theorem to express the number in a closed form? I know Wolfram can't find the antiderivative of x^-x, but does it exist?

Yes, but it's not elementary, which means it won't help us express the number in closed form.
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kg959
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### Re: To what does this series evaluate?

Your sum seems to add smaller and smaller values each time it iterates. This makes me think it might be relatively easily represented with continued fractions.

If you have Maple or something else equally helpful, try running cfrac(1), cfrac(1+(1/4)), cfrac(1+(1/4)+(1/27)) and so on and see if there's a pattern to the coefficients that develop. Your number might be irrational, but it still might be expressible as a pattern of cfrac coefficients.

kg959
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### Re: To what does this series evaluate?

Actually, never mind. I got curious and ran it myself. You function converges to exactly:
$1+(1/(3+(1/4+(1/3+(1/1+(1/2+(1/1+(1/1+(1/6+(1/8+(1/1+(1/55+(1/1+(1/3+(1/1+(1/2+(1/7))))))))))))))))$

The coefficients stop, which means that either Maple is lying about the precision, or your number is rational.

NathanielJ
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### Re: To what does this series evaluate?

kg959 wrote:Actually, never mind. I got curious and ran it myself. You function converges to exactly:
$1+(1/(3+(1/4+(1/3+(1/1+(1/2+(1/1+(1/1+(1/6+(1/8+(1/1+(1/55+(1/1+(1/3+(1/1+(1/2+(1/7))))))))))))))))$

The coefficients stop, which means that either Maple is lying about the precision, or your number is rational.

Maple is almost surely lying to you. I entered that number into WolframAlpha and got 1.08 or something like that. After correcting(?) the parentheses, it still evaluated to 1.3 something, which is larger than the quantity we're interested in.
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skullturf
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### Re: To what does this series evaluate?

I wonder if it's possible to prove that

\sum_{k=1}^{\infty} k^{-k}

is irrational.

An admittedly vague meta-mathematical guideline is that sums that converge "too fast" are often irrational or even transcendental. The denominator k^k is of course pretty large.

There's a simple proof that e is irrational, but it relies on more than just the fact that the denominators in \sum \frac{1}{n!} are "large" -- it relies crucially on the fact that they're factorials.

I believe the first explicit example of a particular number shown to be transcendental was \sum \frac{1}{10^{n!}} (Liouville's constant).

Yakk
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### Re: To what does this series evaluate?

I didn't correct the parenthesis.

I got 1.0879556033621124575214415580707.

I'll agree, the posters parenthesis look off, because they are almost all useless?

=~ 1 + 1/11.369372294372294372294372294371
Or, exactly: = 1 + 1/(11 + 3413/9240)

Note: ~13 bits of precision.
Note note: ln ( N ^ N ) / ln (2) = 1.44 * N * ln(N)
It hits ~13 when crossing 5 to 6.

1/1 + 1/4 + 1/27 + 1/256 + 1/3125 + 1/46656 + 1/823543 + 1/16777216 + 1/387420489 + 1/10000000000 + ...
= 1.2912859970590430507348803843641 + ...

And yes, that is far off.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.