I'm just wondering is there a proof for the sum of the series of general term $\frac{1}{n^n}$. I can't seem to find one online. I know it converges pretty fast but I can't think how you could begin to solve it.
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You mean an explicit formula for $\sum_{n=1}^\infty \frac{1}{n^n}$? – apnorton Jan 29 '14 at 01:15
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IF you want a proof of convergences why not use the root test? – Jose Antonio Jan 29 '14 at 01:16
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Do you want the value of the sum, or just a proof the series converges? If the later, noting that $\frac{1}{n^n} = o\left(\frac{1}{n^2}\right)$ immediately gives it. – Clement C. Jan 29 '14 at 01:17
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I know how to prove convergence but not its sum – James Clarke Jan 29 '14 at 01:18
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I just thought that it would be something elegant like the sum of 1/n^2 because its such a simple series – James Clarke Jan 29 '14 at 01:22
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2possible duplicate of Series as an integral (sophomore's dream) – Nathaniel Bubis Jan 29 '14 at 01:33
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"Proof for the sum" does not mean anything. What precisely are you asking? – Andrés E. Caicedo Jan 29 '14 at 01:46
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There is an amazing result that $$\sum_{n=1}^\infty \frac{1}{n^n}=\int_0^1 \frac{dx}{x^x}\ .$$ This gives you some kind of sum though of course it's not a simple evaluation.
I haven't been able to locate a reference for this result, maybe someone else can help.
David
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that is kinda cool, do you have a proof of the statement I'd love to see it. Haven't a clue how you'd start to evaluate the integral though – James Clarke Jan 29 '14 at 01:25
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Thanks @qwr for providing the name "Sophomore's Dream", looking it up will locate a proof here. – David Jan 29 '14 at 01:32
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It's called Sophomore's Dream, and is equal to
$$ \int_0^1 x^{-x}\ dx $$ around 1.291285997.
qwr
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1That is class, thanks, just googled it, couldn't find it without the name. – James Clarke Jan 29 '14 at 01:34