is equal to :
Question:

$\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots \ldots++\frac{1}{n}}{n^{2}}\right)^{n}$ is equal to :

  1. (1) $\frac{1}{2}$

  2. (2) $\frac{1}{\mathrm{e}}$

  3. (3) 1

  4. (4) 0


Correct Option: , 3

Solution:

It is $1^{\infty}$ form

$\mathrm{L}=\mathrm{e}^{\lim _{z \rightarrow \infty}}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\ldots \ldots+\frac{1}{n}}{n}\right)$

$S=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\ldots \ldots \ldots+\frac{1}{15}\right)$

$S<1+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}\right) \cdots \cdots \cdots+\underbrace{\left(\frac{1}{2^{P}}+\ldots \ldots \ldots+\frac{1}{2^{P}}\right)}_{2^{\mathrm{p}} \text { times }})$

$S<1+1+1+1+\ldots \ldots \ldots+1$

$\$ \mathrm{~S} \therefore \quad L=e^{\lim _{n \rightarrow \infty} \frac{(P+1)}{2^{P}}}$

$\Rightarrow L=e^{\circ}=1$

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