Define a sequence {s_n} of real numbers by s_ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the sequence $s_n = \sum_{k=0}^n \frac{1}{\sqrt{n^2+k}}$.For each $k$ such that $0 \leq k \leq n$,we have $n^2 \leq n^2+k \leq n^2+n$.Taking

Vedclass · Accepted Answer

Exists and lies in the interval $[1, 2)$

Vedclass · Answer

(C) Given the sequence $s_n = \sum_{k=0}^n \frac{1}{\sqrt{n^2+k}}$.For each $k$ such that $0 \leq k \leq n$,we have $n^2 \leq n^2+k \leq n^2+n$.Taking the square root and reciprocal,we get $\frac{1}{\sqrt{n^2+n}} \leq \frac{1}{\sqrt{n^2+k}} \leq \frac{1}{\sqrt{n^2}}$.Summing from $k=0$ to $n$ (which contains $n+1$ terms),we have:$(n+1) \frac{1}{\sqrt{n^2+n}} \leq s_n \leq (n+1) \frac{1}{\sqrt{n^2}}$.As $n \rightarrow \infty$,the left side $\frac{n+1}{\sqrt{n^2+n}} = \frac{n(1+1/n)}{n\sqrt{1+1/n}} = \sqrt{1+1/n} \rightarrow 1$.The right side $\frac{n+1}{n} = 1 + \frac{1}{n} \rightarrow 1$.By the Squeeze Theorem,$\lim_{n \rightarrow \infty} s_n = 1$.Since $1$ lies in the interval $[1, 2)$,the correct option is $C$.

Define a sequence $\{s_n\}$ of real numbers by $s_n = \sum_{k=0}^n \frac{1}{\sqrt{n^2+k}}$,for $n \geq 1$. Then,$\lim_{n \rightarrow \infty} s_n$:

Similar Questions

If $f(x) = \begin{cases} x, & \text{if } x \in \mathbb{Q} \\ -x, & \text{if } x \in \mathbb{Q}^c \end{cases}$,then $\lim_{x \to 0} f(x)$ is

The value of $\lim _{n \rightarrow \infty} \frac{[r]+[2r]+\ldots+[nr]}{n^{2}}$,where $r$ is a non-zero real number and $[x]$ denotes the greatest integer less than or equal to $x$,is equal to:

$\lim_{x \rightarrow \infty} \frac{[2x - 3]}{x} = $

$\mathop {\lim }\limits_{x \to 0} \,\left( {\frac{{x - \sin x}}{x}} \right)\,\sin \left( {\frac{1}{x}} \right)$

If $f(x) = \begin{cases} x\sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module