Find the limit of $\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { \sqrt { { n }^{ 2 } - { k }^{ 2 } } } { { n }^{ 2 } } } } $

Question

Need help to find the limit of $$\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { \sqrt { { n }^{ 2 } - { k }^{ 2 } } } { { n }^{ 2 } } } }$$

Answer 1

Hint:
$$\lim _{ n\rightarrow \infty }{ \sum _{ k=1 }^{ n }{ \frac { \sqrt { { n }^{ 2 } - { k }^{ 2 } } } { { n }^{ 2 } } } }=\lim_{n \to \infty } \sum _{ k=1 }^{ n }\frac{1}{n}\sqrt{1-\frac{k^2}{n^2}}$$

Replace $\frac{1}{n}$ with ${dx}$ and $\frac{k}{n}$ with $x$

$$ \lim _{ n\rightarrow \infty }\sum _{ k=1 }^{ n }\frac{1}{n}\sqrt{1-\frac{k^2}{n^2}}=\int_0^1\sqrt{1-x^2}{dx}$$

I obtained the bounds 0 and 1 using

$$\int_a^bf(x)\,\mathrm{d}x=\lim_{n\to\infty}\sum_k^n\frac{b-a}nf\left(k\frac{b-a}{n}\right)$$

Answer 2

As noticed by Riemann sum we obtain

$$\lim _{ n\rightarrow \infty }\sum _{ k=1 }^{ n }\frac{1}{n}\sqrt{1-\frac{k^2}{n^2}}=\int_0^1\sqrt{1-x^2}{dx}=\frac{\pi}4$$

As an alternative, just to explore other ways, since for a fixed $k$ by binomial series we have

$$\sqrt{1-\frac{k^2}{n^2}}=\sum_{j=0}^\infty \binom{\frac12}{j}\left(-\frac{k^{2}}{n^{2}}\right)^j $$

and therefore by Faulhaber's formula

$$\sum _{ k=1 }^{ n }\frac{1}{n}\sqrt{1-\frac{k^2}{n^2}} =\sum_{j=0}^\infty \left[\binom{\frac12}{j}\frac{(-1)^j}{n^{2j+1}}\sum_{k=1}^n k^{2j}\right] \to\sum_{j=0}^\infty \frac{(-1)^j\binom{\frac12}{j}}{{2j+1}}$$

and since for the binomial coefficient with $n=1/2$ we have

$${{\frac12}\choose{j}}={{2j}\choose{j}}\frac{(-1)^{j+1}}{2^{2j}(2j-1)}\implies \binom{\frac12}{j}\frac{(-1)^j}{{2j+1}}=\frac{\binom{2j}{j}}{2^{2j}(1-4j^2)}$$

we obtain that

$$\sum_{j=0}^\infty \frac{(-1)^j\binom{\frac12}{j}}{{2j+1}}=\sum_{j=0}^\infty \frac{\binom{2j}{j}}{2^{2j}(1-4j^2)}=\frac{\pi}4$$