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

Let $f: R \to R$ be a positive increasing function with $\lim_{x \to \infty} \frac{f(3x)}{f(x)} = 1$. Then $\lim_{x \to \infty} \frac{f(2x)}{f(x)} = $

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

$\lim _{x \rightarrow \infty}\left(\frac{2+\sin x}{x^2+3}\right)$ is equal to

Let $f : (1, 2) \to R$ satisfy the inequality $\frac{\cos(2x - 4) - 33}{2} < f(x) < \frac{x^2 |4x - 8|}{x - 2}$ for all $x \in (1, 2)$. Then $\lim_{x \to 2^-} f(x)$ is equal to

The limit of $x \sin \left(e^{\frac{1}{x}}\right)$ as $x \rightarrow 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module