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 $1,5,32,288,3413,\dots$


3 Answers 3


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

The linked paper is available here

The given bound is

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

  • $\begingroup$ Thank you, does that mean that a closed form is impossible? Or, that none has been found (yet)? $\endgroup$
    – Carolus
    Jun 13, 2011 at 7:52
  • $\begingroup$ @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!) $\endgroup$
    – Juan S
    Jun 13, 2011 at 8:05

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.


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.

  • 8
    $\begingroup$ Then why not $$ n^n\left( 1+\frac{2}{(n-1)}\frac18 \right) \le 1^1 + 2^2 + \cdots + n^n < n^n \left(1+\frac{2}{(n-1)}\frac1e \right) $$ $\endgroup$ Jan 2, 2012 at 15:32

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