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

Solution 1:

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

The linked paper is available here

The given bound is

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

Solution 2:

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.

Solution 3:

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.