Does this expression diverge or converge?

Question

I have the following expression:

$$\lim_{n \to \infty} \frac{2}{n^2} \ {\sum_{i=1}^{n}{\sqrt{n^2 - i^2}}} \ $$

I am not quite sure whether it will converge or diverge. Can somebody tell me how to figure it out?

Answer 1

hint: $S_n = \dfrac{2}{n}\displaystyle \sum_{i=1}^n \sqrt{1-\left(\frac{i}{n}\right)^2}$

Answer 2

Rewriting $$\frac2{n^2}\sum_{i=1}^n\sqrt{n^2-i^2}=\frac1n\sum_{i=1}^n2\sqrt{1-(i/n)^2}, $$ we recognize a Riemann sum for $$\int_0^12\sqrt{1-x^2}\,\mathrm dx.$$

Answer 3

$\sqrt{n^2-i^2} < \sqrt{n^2} $ so $( \frac{2}{n^2} \sum_{i=1}^{n} \sqrt{n^2-i^2} ) < ( \frac{2}{n^2} \sum_{i=1}^{n} \sqrt{n^2} ) = \frac{2}{n^2} n \times n = 2 $ use comparison test. So $ \lim_{n \to \infty} \frac{2}{n^2} \sum_{i=1}^{n} \sqrt{n^2-i^2} < 2 $ so limit must be finite.

Answer 4

The Riemann sum is the most rapid tool for obtain the convergence and the closed form. But if you want another way we can consider the Abel's summation and get $$S_{n}=\sum_{k=1}^{n}\sqrt{n^{2}-k^{2}}=\int_{1}^{n}\frac{\left\lfloor t\right\rfloor t}{\sqrt{n^{2}-t^{2}}}dt $$ where $\left\lfloor t\right\rfloor $ is the floor function. Now, since $t-1\leq\left\lfloor t\right\rfloor \leq t $ we have $$\int_{1}^{n}\frac{t^{2}}{\sqrt{n^{2}-t^{2}}}dt-\sqrt{n^{2}-1}\leq S_{n}\leq\int_{1}^{n}\frac{t^{2}}{\sqrt{n^{2}-t^{2}}}dt $$ and the integral is not difficult to evaluate $$\int_{1}^{n}\frac{t^{2}}{\sqrt{n^{2}-t^{2}}}dt\overset{t=n\sin\left(u\right)}{=}n^{2}\int_{\arcsin\left(1/n\right)}^{\pi/2}\sin^{2}\left(u\right)du $$ $$=\frac{n^{2}}{2}\left(\int_{\arcsin\left(1/n\right)}^{\pi/2}1du-\int_{\arcsin\left(1/n\right)}^{\pi/2}\cos\left(2u\right)du\right) $$ $$=\frac{n^{2}}{2}\left(\frac{\pi}{2}-\arcsin\left(\frac{1}{n}\right)-\frac{1}{n}\cos\left(\arcsin\left(1/n\right)\right)\right) $$ so, by squeeze theorem, we have

$$\frac{2S_{n}}{n^{2}}\rightarrow \color{red}{\frac{\pi}{2}}.$$

Answer 5

Since $f(x)=\sqrt{1-x^2}$ is a concave function on $[0,1]$, the sequence given by

$$ S_n = \frac{1}{n}\sum_{k=1}^{n}\sqrt{1-\left(\frac{k}{n}\right)^2}$$ that is trivially bounded, is also an increasing sequence by the Hermite-Hadamard inequality or by Karamata's inequality. Increasing and bounded implies convergent, hence $$ \lim_{n\to +\infty} S_n = \sup_{n} S_n = \int_{0}^{1}\sqrt{1-x^2}\,dx = \frac{\pi}{4} $$ since the integral is just the area of a quarter circle.