Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

3 Answers

up vote 9 down vote accepted

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

The linked paper is available here

The given bound is

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

share|improve this answer
    
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
add comment

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.

share|improve this answer
1  
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.