What is the limit of given the series?

Question

Find the limit of this series.

$\lim_{n \to \infty} \sum_{i=1}^n {{i \sqrt{n^2 - i^2}} \over n^3}$

Here is my attempt)

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

= $\int_0^1 xi \sqrt{1-(ix)^2}$

= $\int_0 ^1 xi \sqrt{1+x^2}$

= $i$ [${1 \over 3}(1+x^2)^{3\over2} $]$_{0} ^ 1$

= $i \over 3$ $(2 ^ {3 \over 2} -1)$

But the answer was $1 \over 3$. I don't know which point I have a mistake.

Any help would be appreciated.

score 1 · Answer 1 · answered Jan 09 '20 at 12:47

1

As $\sqrt{\dfrac1{n^2}}=\dfrac1n$ for $n>0,$

$$\dfrac{i\sqrt{n^2-i^2}}{n^3}=\dfrac1n\cdot\dfrac in\sqrt{1-\left(\dfrac in\right)^2}$$

Now use The limit of a sum $\sum_{k=1}^n \frac{n}{n^2+k^2}$ and replace $\sqrt{1-x^2}=y\implies1-x^2=y^2$

answered Jan 09 '20 at 12:47

lab bhattacharjee

274,582

Could you please tell me the reason why $i \over n$ is replaced to$ x$ ? Why it is wrong when replacing to $xi$? – se-hyuck yang Jan 09 '20 at 12:56
@se-hyuckyang, Please follow link supplied in the post – lab bhattacharjee Jan 09 '20 at 13:20

score 1 · Accepted Answer · answered Jan 09 '20 at 12:48

1

It's $$\lim_{n\rightarrow+\infty}\sum_{i=1}^n\left(\frac{i}{n}\sqrt{1-\frac{i^2}{n^2}}\frac{1}{n}\right)=\int_0^1x\sqrt{1-x^2}dx=\frac{(1-x^2)^{\frac{3}{2}}}{\frac{3}{2}\cdot(-2)}|_0^1=\frac{1}{3}.$$

answered Jan 09 '20 at 12:48

Michael Rozenberg

194,933

Why does $i \over n$ replaced to $x$? should it be $xi$? – se-hyuck yang Jan 09 '20 at 12:53
@se-hyuck yang I used a definition of the Riemann's integral: https://en.wikipedia.org/wiki/Riemann_integral Draw it! Take $f(x)=x\sqrt{1-x^2}$ on $[0,1]$ and write an integral sum. – Michael Rozenberg Jan 09 '20 at 12:55
Yes it would be.. It looks like a silly thought, But Defining the Riemann's integral,this integral is defined on $[0,1]$ as a partition $ 1 \over n $ not the $[0,i]$(partition $ i \over n $) doesn't it? – se-hyuck yang Jan 09 '20 at 13:01
@se-hyuck yang You need to take $\xi_i\in\left[\frac{i-1}{n},\frac{i}{n}\right]$ and we can take $\xi_i=\frac{i}{n}.$ – Michael Rozenberg Jan 09 '20 at 13:04
Ah.. I had a mistake that the regarding the $i$ as a imaginary number not the index of sum. Thanks for What I've missed. :) – se-hyuck yang Jan 09 '20 at 13:26
@se-hyuck yang You are welcome! – Michael Rozenberg Jan 09 '20 at 13:27

score 0 · Answer 3 · answered Jan 10 '20 at 14:31

If $b=1,a=0$

$$\int_{0}^{1}f(x)dx=\lim_{n\to\infty}\dfrac{1}{n}\sum_{i=1}^{n}f\left(\dfrac{i}{n}\right)\tag{1}$$

So to convert this limit $\lim_{n\to\infty}\sum_{i=1}^{n}\dfrac{i\sqrt{n^2-i^2}}{n^3}$ in the form of equation $(1)$, we rewrite it as follows:-

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

So we can say $f\left(\dfrac{i}{n}\right)=\dfrac{i}{n}\sqrt{1-\dfrac{i^2}{n^2}}$

Replacing $\dfrac{i}{n}$ with $x$

$$f(x)=x\sqrt{1-x^2}$$

$$\int_{0}^{1}f(x)dx=\int_{0}^{1}x\sqrt{1-x^2}dx$$

$$1-x^2=t$$ $$-2x=\dfrac{dt}{dx}$$ $$xdx=-\dfrac{dt}{2}$$

$$\int_{0}^{1}f(x)dx=\dfrac{1}{2}\cdot\int_{0}^{1}t^\frac{1}{2}dt$$

$$\int_{0}^{1}f(x)dx=\dfrac{1}{3}$$

Hope it will help you in understanding why we integrals are helpful in such cases.

What is the limit of given the series?

3 Answers3