Does $\sum_{k=0}^{n-1}\frac{1}{\sqrt{n^2-k^2}}\to \int_0^{1-}\frac{1}{\sqrt{1-x^2}}$? Consider the sum $$\sum_{k=0}^{n-1}\frac{1}{\sqrt{n^2-k^2}}.$$
Obviously, 
$$\sum_{k=0}^{n-1}\frac{1}{\sqrt{n^2-k^2}}=\frac{1}{n}\sum_{k=0}^{n-1}\frac{1}{\sqrt{1-\frac{k^2}{n^2}}}.$$
I would like to say that it converges to $\int_0^1\frac{1}{\sqrt{1-x^2}}dx$, but since $f(x)=\frac{1}{\sqrt{1-x^2}}$ is defined and continuous on $[0,1)$ and not on $[0,1]$, I'm not sure if my guess is really true. How can I be really sure about that ?

The integrand is unbounded and, hence, not Riemann integrable.  However the improper integral over $[0,1)$  is convergent (to $\pi/2$).
In general, Riemann sums may not converge for an improper integral. In this case, using the left endpoints the sums do converge since the integrand is monotone.
See here for a justification.
