# Can I get better approximation of $\sum_{k=1}^{n} k^k$

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? –  Pedro Tamaroff Sep 8 '12 at 1:46
@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. –  Pedro Tamaroff Sep 8 '12 at 1:56
@PeterTamaroff Yeah,Sorry for my mistake –  Golbez Sep 8 '12 at 2:24