# Closed form formula for $\sum\limits_{k=1}^n k^k$

Is there a way of finding a formula for $\sum\limits_{k=1}^n k^k$? Maybe I'm missing something really obvious, but I've looked around a bit on the Internet and I haven't been able to find anything.

So, what I'm looking for is a formula in closed form to generate the sequence $1,5,32,288,3413,\dots$

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Have a look on OEIS - it would appear there is no simple closed form.

The linked paper is available here

The given bound is

$$n^n\left( \frac{4n-3}{4n-4} \right) \le 1^1 + 2^2 + \cdots + n^n < n^n \left(\frac{2+e(n-1)}{e(n-1)}\right)$$

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Thank you, does that mean that a closed form is impossible? Or, that none has been found (yet)? – Carolus Jun 13 '11 at 7:52
@Carolus - if by 'closed form' you mean in terms of 'elementary functions' I would say it is highly unlikely such a form exists (but do see Wikipedia as closed form is never really explicit!) – Juan S Jun 13 '11 at 8:05
thanks again! – Carolus Jun 13 '11 at 8:32

The OEIS doesn't list a closed form for this sequence, only noting that $a_{n+1}/a_n>en$, and $a_{n+1}/a_n\to en$ as $n\to\infty$. There's also a list of the first 100 values of the sequence here.

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I would write the inequality as

$$n^n\left( 1+\frac{1}{4(n-1)} \right) \le 1^1 + 2^2 + \cdots + n^n < n^n \left(1+\frac{2}{e(n-1)}\right)$$

to better show the bounds.

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Then why not $$n^n\left( 1+\frac{2}{(n-1)}\frac18 \right) \le 1^1 + 2^2 + \cdots + n^n < n^n \left(1+\frac{2}{(n-1)}\frac1e \right)$$ – Gottfried Helms Jan 2 '12 at 15:32