Question

Let $nϵN$ & the A.M., G.M., H.M. & the root mean square of $n$ numbers $2n+1,2n+2,...,$ up to $n^{th}$ number are $A_n$, $G_n$, $H_n$ and $R_n$ respectively.
On the basis of above information answer the following questions$\underset{n\to \infty }{lim}\frac{n}{H_n}$ equals

• A. $\log \left(\frac{2}{3}\right)$
• B. $\log 6$
• C. $\log \left(\frac{3}{2}\right)$
• D. None of these

Answer 1

The correct option is C $\log \left(\frac{3}{2}\right)$

$H_n=$ H.M. of $2n+1,2n+2,....,2n+n$

$H_n=\frac{n}{\frac{1}{2n+1}+\frac{1}{2n+2}+...+\frac{1}{2n+n}}$

$\therefore$ $\frac{1}{H_n}=\frac{\frac{1}{2n+1}+\frac{1}{2n+2}+...+\frac{1}{2n+n}}{n}$

$\Rightarrow$ $\frac{n}{H_n}=\frac{1}{n}[\frac{1}{\frac{2n+1}{n}}+\frac{1}{\frac{2n+2}{n}}+...+\frac{1}{\frac{2n+n}{n}}]=\frac{1}{n}\sum _{r=1}^n\left(\frac{1}{2+\frac{r}{n}}\right)$

$\therefore$ $\underset{n\to \infty }{lim}\frac{n}{H_n}=\sum _{r=1}^n\frac{1}{n}\left(\frac{1}{2+\frac{r}{n}}\right)$

$={\int }_0^1\frac{1}{2+x}dx={\left(\log (2+x)\right)}_0^1=\log \left(\frac{3}{2}\right)$
Ans: C