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

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 $$\infty$$?

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