_ n (1+ 1+1 2 + .+1 n n^ 2 )^ n =......... - Vidyadip

MCQ

$\lim _{n \rightarrow \infty}\left(1+\frac{1+\frac{1}{2}+\ldots \ldots .+\frac{1}{n}}{n^{2}}\right)^{n}=.........$

A
$\frac{1}{2}$
B
$0$
C
$\frac{1}{ e }$
✓
$1$

✓

Answer

Correct option: D.

$1$

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

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

Now,

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

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

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

$=1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The area bounded by the $x - $ axis and the curve $y = \sin x$ and $x = 0,$ $x = \pi $ is

Let $\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}$. Let $\hat{\mathrm{c}}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$. Then such a vector $\hat{c}$ is :

The area of a parallelogram whose adjacent sides are given by the vectors $i + 2j + 3k$ and $ - 3i - 2j + k$ (in square unit) is

Let $\mathrm{z}$ be a complex number such that $|\mathrm{z}+2|=1$ and $\operatorname{Im}\left(\frac{z+1}{z+2}\right)=\frac{1}{5}$. Then the value of $|\operatorname{Re}(\overline{z+2})|$ is :

If $\alpha, \beta$ are the roots of the equation, $x^2-x-1=0$ and $S_n=2023 \alpha^n+2024 \beta^n,$ then

If ${ }^{2 n+1} P_{n-1}:{ }^{2 n-1} P_n=11: 21$, then $n^2+n+15$ is equal to:

$\int\limits_0^{\frac{\pi }{4}} {} (tan^n x + tan^{n -2} x) d (x - [x])$ is : ( $[. ]$ denotes greatest integer function)

If $x + y - z = 0,\,3x - \alpha y - 3z = 0,\,\,x - 3y + z = 0$ has non zero solution, then $\alpha = $

If $\mathrm{x}=2 \sin \theta-\sin 2 \theta$ and $\mathrm{y}=2 \cos \theta-\cos 2 \theta$ ; $\theta \in[0,2 \pi],$ then $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ at $\theta=\pi$ is :

If $y = 2sin^{-1} \sqrt {(1-x)} + sin^{-1} (2\sqrt{x(1-x)})$ for $0 < x < \frac{1}{2}$, then $\frac{dy}{dx}$ equals-