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  • <img alt="\mathrm{\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots .+\frac{1}{n}}{n^2}\right)^n}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bn%20%5C

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

Option: 1

1


Option: 2

3


Option: 3

2


Option: 4

0


Answers (1)

best_answer

Given limit is of \mathrm{1^{\infty}} form.
So, given limit \mathrm{(l)=\exp \left(\lim _{n \rightarrow \infty} \frac{1+\frac{1}{2}+\frac{1}{3}+\ldots .+\frac{1}{n}}{n}\right)}
Now, \mathrm{0 \leq 1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} \leq 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots .+\frac{1}{\sqrt{n}}}

                                                                                               \mathrm{ \leq 2 \sqrt{n}-1 }
\mathrm{ So, l=\exp (0) }                                                       (From Sandwich theorem)

           \mathrm{ =1 }

Posted by

Devendra Khairwa

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