Prove the following

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. $\frac{1}{2}$

  2.  0

  3. $\frac{1}{\mathrm{e}}$

  4. 1


Correct Option: , 4

Solution:

Given limit is of $1^{\infty}$ form

So, $l=\exp \left(\lim _{n \rightarrow \infty} \frac{1+\frac{1}{2}+\frac{1}{3}+\ldots \ldots . .+\frac{1}{n}}{n}\right)$

Now,

$0 \leq 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{\mathrm{n}} \leq 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots .+\frac{1}{\sqrt{\mathrm{n}}}$

$\leq 2 \sqrt{n}-1$

So, $l=\exp (0)$ (from sandwich theorem)

$=1$

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