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