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I stumbled across this problem to find the result of the following expression: $$\sum_{n=1}^\infty \frac{1}{n^n}$$

but I don't know how to approach it. It was suggested to me that I try this: $$\sum_{n=1}^\infty e^{-n\ln n}$$ However, I still don't know what to do. How can I find a closed form?

Michael Hardy
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aaazalea
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  • do you need the convergence? than compare to $n!$ or $n^2$ or something like that – Dominic Michaelis Mar 13 '13 at 17:59
  • there is no known closed form, but there is an integral with the same value. –  Mar 13 '13 at 17:59
  • I want to find a closed form for the sum. – aaazalea Mar 13 '13 at 17:59
  • There is no closed form and this has been asked before, by me, for example, and it was closed as a duplicate. – Valtteri Mar 13 '13 at 17:59
  • Could you close the question as a duplicate of that then? and what is the integral? – aaazalea Mar 13 '13 at 18:00
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    http://en.wikipedia.org/wiki/Sophomore%27s_dream and I have no rights to close anything :D – Valtteri Mar 13 '13 at 18:01
  • flag it I mean. I'll close it if I get the link. – aaazalea Mar 13 '13 at 18:03
  • See my comment above for the link to a previous version to the question. – Cameron Buie Mar 13 '13 at 18:07
  • @Jake223 : Please don't write n\ ln\ n. Instead write n\ln n. The backslash on \ln not only prevent it from being italicized as if the letters were variables, but also results in proper spacing. TeX is not the work of primitive cave men; it is sophisticated. (I edited the question accordingly.) – Michael Hardy Mar 13 '13 at 18:10
  • While we are here, what are the chances to find $\displaystyle \sum_{n=1}^{\infty} n! \tan \left(\frac{1}{n^n}\right)$? – J.H. Mar 13 '13 at 18:13
  • in terms of integral i think i saw this sum and was equal to $$ \int_{0}^{1}dxx^{-x}= \sum_{1}^{\infty}n^{-n} $$ – Jose Garcia Mar 13 '13 at 18:31
  • I would like to mention that I am (with @Marvis) (though much delayed) working on a paper that generalizes the sophomore's dream series and integrals as a kind of deformed exponential. – graveolensa Mar 13 '13 at 18:32

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