What will be the general formula for the series: $1^1 + 2^2 + 3^3 + 4^4\ldots$
The series diverges, but can we have a formula to find the sum of $n$ terms.
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$\begingroup$ There is no known general formula. $\endgroup$– Simply Beautiful ArtCommented Nov 10, 2016 at 15:15
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1$\begingroup$ It's: $\sum\limits_{n=1}^{+\infty} n^n=+\infty$ $\endgroup$– MohamedCommented Nov 10, 2016 at 15:16
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$\begingroup$ I do understand that the series diverges, but can we get a formula to calculate the sum? $\endgroup$– Harsha G.Commented Nov 10, 2016 at 15:17
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$\begingroup$ @Mohamed The title suggests finite sum, but the question and tags suggestion infinite sum. Please clarify. $\endgroup$– Simply Beautiful ArtCommented Nov 10, 2016 at 15:18
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$\begingroup$ @iSimple Art: Thank's. I was attracted by dots. $\endgroup$– MohamedCommented Nov 10, 2016 at 15:21
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1 Answer
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The function $f(n)=\sum_{k=1}^n k^k$ is called the hypertriangular function. There are some papers on it, for example
Mohammad K. Azarian, On the hyperfactorial function, hypertriangular function and the discriminants of certain polynomials. Int. J. Pure Appl. Math. 36 (2007), 251-257.
There some estimates are shown, like $$ n^n\frac{4n-3}{4n-4}<1^1+2^2+3^3+\cdots +n^n<n^n\frac{2+e(n-1)}{e(n-1)} $$ for all $n>1$.
The problem to find a formula is a problem by G.W. Wishard, posed in $1945$, see here.
answered Nov 10, 2016 at 15:21
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$\begingroup$ Hint, better bounds by using the fact that $f(n)\in\mathbb N$ $\endgroup$ Commented Nov 10, 2016 at 15:24