Question

$\underset{n\to \infty }{lim}{(1+\frac{1+\frac{1}{2}+......+\frac{1}{n}}{n^2})}^n$ is equal to :

• A. $\frac{1}{2}$
• B. $\frac{1}{e}$
• C. $1$
• D. $0$

Answer 1

The correct option is C $1$

It is $1^{\infty }$ form
$L=e^{\underset{n\to \infty }{lim}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}}{n}\right)}S=1+(\frac{1}{2}+\frac{1}{3})+(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7})+(\frac{1}{8}+....+\frac{1}{15})S<1+(\frac{1}{2}+\frac{1}{2})+(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4})......+\underset{2^P\text{times}}{(\frac{1}{2^P}+.....+\frac{1}{2^P})}S<1+1+1+1+....+1S<P+1\therefore L=e^{\underset{n\to \infty }{lim}\frac{(p+1)}{2^{P+1}-1}}\Rightarrow L=e^{\circ }=1$