TITLE: A very different alternative form of the geometric series
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: $$\lim_{n\to \infty} \sum_{k=1}^n \frac{1}{\sqrt[x]{n^{x-1}k}} = $$ 
$$\lim_{n\to \infty}\sum_{k=1}^n (n^{x-1}k)^{-\frac{1}{x}} = $$
$$\lim_{n\to \infty}\frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^{-\frac{1}{x}} = $$
$$\int_{0}^1t^{-\frac{1}{x}}dt = \frac{x}{x-1} $$