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Is it possible to get approximation$f(n)$ of $\sum_{k=1}^{n} k^k$ with \begin{align} \lim_{n\to \infty }\left(f(n)-\sum_{k=1}^{n} k^k\right)=0 \end{align} Thanks for your attention!

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$\sum_{k=1}nn^n=n\cdot n^n=n^{n+1}$; is that really what you intended? – Brian M. Scott Sep 8 '12 at 1:46
You mean $$\sum_{k=1}^{n} k^k$$ right? – Peter Tamaroff Sep 8 '12 at 1:46
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@Peter: I don’t think that it’s appropriate to change the substance of a question without the OP’s input. – Brian M. Scott Sep 8 '12 at 1:49
@BrianM.Scott Oh, well. – Peter Tamaroff Sep 8 '12 at 1:56
@PeterTamaroff Yeah,Sorry for my mistake – Golbez Sep 8 '12 at 2:24
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