Question

For non-zero distinct constants $a$ and $b,$ the value of $\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{\sqrt{n}}{\sqrt{r}{(a\sqrt{n}-b\sqrt{r})}^2}$ is

• A. $\frac{2}{a(a-b)}$
• B. $\frac{3}{a(a-b)}$
• C. $\frac{2}{a(b-a)}$
• D. $\frac{3}{a(b-a)}$

Answer 1

The correct option is A $\frac{2}{a(a-b)}$

We have $I=\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{\sqrt{n}}{\sqrt{r}{(a\sqrt{n}-b\sqrt{r})}^2}$
$=\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=1}^n\frac{1}{\sqrt{\frac{r}{n}}{(a-b\sqrt{\frac{r}{n}})}^2}=\underset{0}{\overset{1}{\int }}\frac{dx}{\sqrt{x}{(a-b\sqrt{x})}^2}$
Putting $(a-b\sqrt{x})=t\Rightarrow \frac{dx}{\sqrt{x}}=\frac{2dt}{-b}$
When $x\to 0,(a-b\sqrt{x})\to a$
When $x\to 1,(a-b\sqrt{x})\to a-b$
$\therefore I=\frac{-2}{b}\underset{a}{\overset{a-b}{\int }}\frac{dt}{t^2}$
$=\frac{2}{b}{\left[\frac{1}{t}\right]}_a^{a-b}=\frac{2}{b}(\frac{1}{a-b}-\frac{1}{a})=\frac{2}{a(a-b)}$