Question

Let $n\in N$ and $n+1,n+2,n+3,,,,,n+n$ are $n$ numbers if H.M be the harmonic mean of these $n$ numbers then ${lim}_{n\to \infty }\frac{H_n}{n}$ equals

• A. $\log \sqrt{2}$
• B. $\log \sqrt{3}$
• C. ${\left(\log 2\right)}^{-1}$
• D. None of these

Answer 1

The correct option is C ${\left(\log 2\right)}^{-1}$

$H_n=\frac{n}{\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+....+\frac{1}{n+n}}$
$\frac{n}{H_n}={\sum }_{r=1}^n\frac{1}{n+r}\frac{n}{H_n}={\sum }_{r=1}^n\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right)$
$\therefore {lim}_{n\to \infty }\frac{n}{H_n}={lim}_{n\to \infty }{\sum }_{r=1}^n\frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right)$
$={\int }_0^1\frac{1}{1+x}dx=\log 2$
$\therefore {lim}_{n\to \infty }\frac{n}{H_n}={\left(\log 2\right)}^{-1}$