I was thinking about this series: $$ \sum_{k=1}^{\infty}\left( \frac{1}{k}\right)^k$$ I calculated the limit $$\lim_{k \rightarrow \infty} \left(\frac{1}{k}\right)^k$$ and the result was $0$, so i got the necessary condition, but it's not sufficient. I can't think in other ways to prove this converges( altough i'm almost shure it does) and more important than this, calculate this sum. Thanks in advance.
Prove convergence and calculate sum (1/k)^k
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sequences-and-series
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0Hint for convergence: For $k \ge 2$, we have $(1/k)^k \le (1/k)^2$. – 2017-01-11
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1It can be shown that the value of the sum is equal to $\int_0^1\frac{dx}{x^x}$ by rewriting $x^x=\exp(x\log x)$, expanding the exponential function as a power series and integrating termwise, but I doubt that there is a closed form for the value of the sum. It is approximately equal to 1.29129. – 2017-01-11
1 Answers
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First of all let's consider sum of $\displaystyle \frac{\pi^2}{6} = \sum_{k = 1}^{\infty}\frac{1}{k^2} > \sum_{k = 1}^{\infty}\frac{1}{k^k}$, so by comparasion test this serie is converge.
What about sum result ? It's better to read this Sophomore's dream.