Does $\sum_{k=0}^{n-1}\frac{1}{\sqrt{n^2-k^2}}\to \int_0^{1-}\frac{1}{\sqrt{1-x^2}}$?

Question

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 ?

Answer 1

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.