Determine if limit of this sequence exists and calculate the limit if it's possible

Question

I have such an sequence: $a_{n} = (1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n})\log (\frac{n+1}{n})$

Could you tell me if I can divide this in two sequences to determine, if they have a limit?

$b_{n} = (1+\sqrt{2}+\sqrt[3]{3}+...+\sqrt[n]{n})$

$b_{n}$ is not bounded above and $b_{n} \to \infty$: exists $c_{n} \le b_{n} : c_{n} \to \infty$

We can take $c_{n} = n\sqrt[n]{n} \to \infty$

$c_{n} = \log (\frac{n+1}{n}) = \log(1 + \frac{1}{n})^n)^\frac{1}{n}) = \frac{1}{n}loge$

I'm using Stolt'z theorem:

$x_{n} = loge$, $y_{n} = n, y_{n} \to \infty$ and it's strictly monotnic, so:

$\frac{x_{n+1}-x_{n}}{y_{n+1}-y_{n}} = \frac{loge - loge \to 0}{n+1-n\to1} = 0$, so $\frac{loge}{n} \to 0$

But now I get lost, how can I mix up $0$ and $\infty$?

Answer 1

As $n \to \infty$, $\log \left( \frac{n+1}{n} \right) \simeq \frac{1}{n}$. Hence, by Cesaro theorem, $$ \lim_{n \to \infty} \log \left( \frac{n+1}{n} \right) \sum_{k = 1}^n \sqrt[k]{k} = \lim_{n \to \infty} \frac{1}{n} \sum_{k = 1}^n \sqrt[k]{k} = \lim_{n \to \infty} \sqrt[n]{n} = 1. $$