Evaluate:

        n + n2 + n3 +...  ...+ nn
Limit  --------------------------
n→∞    1n + 2n + 3n +...  ...+ nn

From my previous comment:
The numerator should be:
n * (n^n - 1)/(n-1)

Evaluating the denominator for different values of n leads to Sloane's oeis A031971.

And looking through the notes, there are several approximations for the value of this series.

One is:  a(n) is asymptotic to (e/(e - 1))*n^n. - Benoit Cloitre, Dec 17 2003

So taking the numerator as:    n * (n^n - 1)/(n-1)
and taking the denominator as: (e/(e - 1))*n^n

Cancel out n/(n-1)
Cancel out (n^n - 1)/n^n

And we have (e - 1)/e =~ 0.632120558828558

Which is close to my f(10,000) = 0.6322235834032078

Posted by Larry on 2023-05-27 14:35:27