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I like to find a formula for the $\sum_{i=1}^n\,i^i$.

It is possible to write a formula for the summation of the form $\sum_{i=1}^n\, i^a$. For example, here explain its methods completely. But what kind of method should we use when the power in the summation is not fixed?

I first tried to find an upper bound. And I could only think of $\sum_{i=1}^n\,i^i\leq n.n^n$ which seems that it's not very useful.

I also wish to compute $n^n=c$, for some constant $c\in \mathbb{N}$. But again it seems very hard. If the power was fixed (that is n^a=c) then we could find the $a$th root of $n$. We could use logarithm if we had the form $a^n=c$ for some fixed $a$. Unfortunately, neither power nor base is fixed.

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  • $\begingroup$ This may help you: Sum of Powers $\endgroup$ – user507623 Dec 3 '17 at 7:57
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This is looking as the somophore's dream.

As far as I can remember, Underwood proposed bounds for this summation.

$$n^n\left(1+\frac 1{4(n-1)} \right) <\sum_{i=1}^{n}\, i^i <n^n\left(1+\frac 2{e(n-1)} \right)$$

Let us check $$\left( \begin{array}{cccc} n & \text{lower} &\text{exact} &\text{upper}\\ 2 & 5. & 5 & 6.94304 \\ 3 & 30.375 & 32 & 36.9327 \\ 4 & 277.333 & 288 & 318.785 \\ 5 & 3320.31 & 3413 & 3699.81 \\ 6 & 48988.8 & 50069 & 53521.5 \\ 7 & 857857. & 873612 & 924531. \\ 8 & 1.73764\times 10^7 & 17650828 & 1.85406\times 10^7 \\ 9 & 3.99527\times 10^8 & 405071317 & 4.23051\times 10^8 \\ 10 & 1.02778\times 10^{10} & 10405071317 & 1.08175\times 10^{10}\\ 11 & 2.92444\times 10^{11} & 295716741928 & 3.06304\times 10^{11} \\ 12 & 9.11874\times 10^{12} & 9211817190184 & 9.51247\times 10^{12} \\ 13 & 3.09185\times 10^{14} & 312086923782437 & 3.21445\times 10^{14} \\ 14 & 1.13257\times 10^{16} & 11424093749340453 & 1.17409\times 10^{16} \\ 15 & 4.45713\times 10^{17} & 449317984130199828 & 4.60907\times 10^{17} \end{array} \right)$$

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  • $\begingroup$ Does this summation have any special name? Would you please tell me where can I find the reference to this proposed bound? $\endgroup$ – Doralisa Dec 3 '17 at 9:05

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