Evaluate the limit of a series with an integral

[e^(i Pi)+1=0]

Homework Statement

\lim_{n \to \infty} \sum_{i=1}^{n} \frac{4}{n}\sqrt{\frac{4i}{n}}

The Attempt at a Solution

This seems to give the right answer, 16/3, but I can't figure out why:

\lim_{n \to \infty}\int_{1}^{n}\frac{4}{n}\sqrt{\frac{4x}{n}}dx

 

[vanhees71]

I'd rather think that you have to interpret the sum as an approximation to the (Riemann) integral
I=\int_0^{4} \mathrm{d} x \sqrt{x},
where the interval is divided in n subintervals of equal size.

 

[HallsofIvy]

This is more than just approximating. If you have the function f(x)= \sqrt{x}, on the interval [0, 4], the Riemann sum, dividing [0, 4] into n equal subintervals, so that each subinterval has length 4/n and x= 4i/n, gives \sum{i= 1}^n \frac{4}{n}\sqrt{\frac{4i}{n}}. As n goes to infinity, we are dividing the interval into more and more smaller and smaller intervals and the limit is the Riemann integral.

 

[haruspex]

This is more than just approximating.

vanhees71 said the sum approximates the Riemann integral; you're saying the limit of the sum is the Riemann integral. No contradiction there.