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

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(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

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

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(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,$\lim_{n \rightarrow \infty} s_n$:

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