A very different alternative form of the geometric series

Question

While I was playing around with series that came up in a calculus assignment I came across this thing here: $$\lim_{n\to \infty} \sum_{i=1}^n \frac{1}{\sqrt[x]{n^{x-1}i}}$$ And after using wolframalpha to evaluate it at some positive real numbers $x$ it seems that $$\lim_{n\to \infty} \sum_{i=1}^n \frac{1}{\sqrt[x]{n^{x-1}i}} = \frac{x}{x-1}$$ which is the value of the geometric series with parameter $\frac{1}{x}$. Unfortunately I have no clue how to prove this.

Answer 1

$$\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{\sqrt[x]{n^{x-1}k}} = $$

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

$$\int_{0}^1t^{-\frac{1}{x}}dt = \frac{x}{x-1} $$