Limit of average of $n^{1/k}$

Define $S_n$ as follows

$$S_n =   \sum_{k=1}^{n} n^{\frac{1}{k}}$$

For eg 

$$S_{10} = 10 + 10^{1/2} + 10^{1/3} + \dots + 10^{1/10} \approx 25.4211$$

Find

$$\displaystyle \lim_{n \to \infty} \dfrac{S_n}{n}$$  

Scroll down for a solution.

We will solve this using the arithmetic mean geometric mean inequality!

For $k \ge 2$ let $$x_1 = x_2 = \dots = x_{k-2} = 1, x_{k-1} = x_k = \sqrt{n}$$

Applying AM GM to these we get 

$$\frac{k-2 + 2\sqrt{n}}{k} \ge n^{1/k} \ge 1$$

Thus 

$$1 - \frac{2}{k} + 2 \frac{\sqrt{n}}{k} \ge n^{1/k} \ge 1$$

Now $\sum_{k=2}^{n} \frac{1}{k} = \log n + O(1)$

Thus

$$n-1 - 2(\log n + O(1)) + 2\sqrt{n}(\log n + O(1)) \ge \sum_{k=2}^n n^{1/k} \ge n-1$$  

And so 

$$2n-1 - 2(\log n + O(1)) + 2\sqrt{n}(\log n + O(1)) \ge \sum_{k=1}^n n^{1/k} \ge 2n-1$$   

Thus 

$$2 + O\left(\frac{\log n}{\sqrt{n}}\right) \ge \frac{S_n}{n} \ge 2 + O\left(\frac{1}{n}\right)$$

Thus $$\frac{S_n}{n} \to 2$$