Question

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

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

Answer 1

The correct option is C

$1$

Explanation for the correct option:

Evaluating $\underset{n\to \infty }{\lim }{\left(1+\frac{1+{\displaystyle \frac{1}{2}}+.....+{\displaystyle \frac{1}{n}}}{n^2}\right)}^n$

We can see that after putting $n=\infty$

We get $1^{\infty }$ form

Considering,

$\Rightarrow L={e^{\underset{n\to \infty }{\lim }\left(1+\frac{1+{\displaystyle \frac{1}{2}}+.....+{\displaystyle \frac{1}{n}}}{n^2}\right)}}^n$

Again consider,

$$\begin{gathered} S=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+.....+\frac{1}{15}\right)+... \\ S<1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right)+.......\left(\frac{1}{2^P}+.....+\frac{1}{2^P}\right) \\ S<1+1+1+.........+1 \\ S<P+1 \end{gathered}$$

Therefore,

$$\begin{gathered} \Rightarrow L=e^{\underset{n\to \infty }{\lim }\frac{\left(p+1\right)}{2^{p+1}-1}} \\ =e^0 \\ =1 \end{gathered}$$

Hence, option C is the correct answer.