Krdytkyu

$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)$$

My try:

The limit can be written as follows:

$$lim_{ntoinfty}left(frac{1}{n^2}cdotsum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}right)$$

Evaluate the following series:

$sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$

$frac{(k+1)^{k}}{k^{k-1}}=kcdotfrac{(k+1)^{k}}{k^{k}}=kcdotleft(1+frac{k+1}{k}-1right)^k=kcdotleft(1+frac{1}{k}right)^k$

Then:

$lim_{ktoinfty}frac{(k+1)^{k}}{k^{k-1}}=lim_{ktoinfty}kcdotleft(1+frac{1}{k}right)^k=ecdotinftyneq0 Longrightarrow sum_{k=1}^{infty}frac{(k+1)^{k}}{k^{k-1}}$ diverges.

Therefore:

$$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=0cdotinfty$$

What to do next?

HINT

We have that by Stolz-Cesaro

$$frac{a_n}{b_n}=frac{sum_{k=1}^{n}frac{(k+1)^{k}}{k^{k-1}}}{n^2}$$

$$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^{n}}}{(n+1)^2-n^2}=frac{(n+2)^{n+1}}{(2n+1)(n+1)^{n}}=frac{n+1}{2n+1}left(frac{n+2}{n+1}right)^{n+1}$$

Hint: $$frac{(k+1)^k}{k^{k-1}} = k left(1+frac{1}{k}right)^k$$
and
$$ left(1+frac{1}{k}right)^k = expleft(k lnleft(1+frac{1}{k}right)right) = expleft(1 + O(1/k)right) = e + O(1/k)$$
Now, what can you say about $$sum_{k=1}^n k (e + O(1/k))$$
?

Thanks to user xbh for the hint:

Using the Stolz-Cesàro theorem, we have:

$a_n=frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}$

$b_n=n^2$

Two monotone and increasing sequences.

Apply the theorem to get:

$frac{a_{n+1}-a_n}{b_{n+1}-b_n}=frac{frac{(n+2)^{n+1}}{(n+1)^n}}{2n+1}=frac{(n+2)(n+2)^{n}}{(2n+1)(n+1)^n}=frac{n+2}{2n+1}cdotleft(1+frac{n+2}{n+1}-1right)^n=frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}$

Then:

$lim_{ntoinfty}frac{1}{n^2}left(frac{2}{1}+frac{9}{2}+frac{64}{9}+cdots+frac{(n+1)^{n}}{n^{n-1}}right)=lim_{ntoinfty}frac{n+2}{2n+1}cdotleft[left(1+frac{1}{n+1}right)^{n+1}right]^{frac{1}{n+1}n}=frac{e}{2}$