Evaluate the limit of $\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)$

Question

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

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My try:

The limit can be written as follows:

$$\lim_{n\to\infty}\left(\frac{1}{n^2}\cdot\sum_{k=1}^{n}\frac{(k+1)^{k}}{k^{k-1}}\right)$$

Evaluate the following series:

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

$\frac{(k+1)^{k}}{k^{k-1}}=k\cdot\frac{(k+1)^{k}}{k^{k}}=k\cdot\left(1+\frac{k+1}{k}-1\right)^k=k\cdot\left(1+\frac{1}{k}\right)^k$

Then:

$\lim_{k\to\infty}\frac{(k+1)^{k}}{k^{k-1}}=\lim_{k\to\infty}k\cdot\left(1+\frac{1}{k}\right)^k=e\cdot\infty\neq0 \Longrightarrow \sum_{k=1}^{\infty}\frac{(k+1)^{k}}{k^{k-1}}$ diverges.

Therefore:

$$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)=0\cdot\infty$$

What to do next?

Answer 1

We have that by Stolz-Cesaro

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

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

Answer 2

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

Answer 3

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:

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

Then:

$\lim_{n\to\infty}\frac{1}{n^2}\left(\frac{2}{1}+\frac{9}{2}+\frac{64}{9}+\cdots+\frac{(n+1)^{n}}{n^{n-1}}\right)=\lim_{n\to\infty}\frac{n+2}{2n+1}\cdot\left[\left(1+\frac{1}{n+1}\right)^{n+1}\right]^{\frac{1}{n+1}n}=\frac{e}{2}$