IF $\displaystyle f(x)=\frac{x}{\sqrt{x+1}}$,
Calculate
$\displaystyle \underset{n \to \infty }{lim}\left [f\left ( \frac{1}{n^2}\right ) +f\left ( \frac{2}{n^2}\right ) + ... +f\left ( \frac{n}{n^2}\right ) \right ]$
Here is an illustration.
$\displaystyle \displaystyle \underset{n \to \infty }{\lim}\left [f\left ( \frac{1}{n^2}\right ) +f\left ( \frac{2}{n^2}\right ) + ... +f\left ( \frac{n}{n^2}\right ) \right ] = \underset{n \to \infty }{\lim}\sum_{k=1}^n f\left ( \frac{k}{n^2}\right ) = $
[Math]\displaystyle = \underset{n \to \infty }{\lim}\sum_{k=1}^n \frac{k/n^2}{\sqrt{k/n^2+1}} [/tex]
which I'm doubtful that you can relate to a Riemann sum, since it doesn't sum to n^2 but to n (so the interval of summation gets smaller and smaller every time we increase n) and we are missing the multiplication by the length of the subdivision 1/n^2. This is in fact why it converges because if one is present without the other we either get divergence or convergence to 0.