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Determine whether the sequence $\{a n\}$ is convergent or divergent, if it is convergent Find the lim

$a n=\frac{\sqrt{1+\tan ^{-1}(n)}-\sqrt{1+\sin (n)}}{n^{2}}$

$\lim _{n \rightarrow \infty}\{a n\}=\lim \left[\frac{\sqrt{1+\tan ^{-1}(n)}-\sqrt{1+\sin (n)}}{n^{2}}\right]$

$=\lim _{n \rightarrow \infty}\left[\frac{\sqrt{1+\tan ^{-1}(n)}}{n^{2}}-\frac{\sqrt{1+\sin (n)}}{n^{2}}\right]$

$\lim _{n \rightarrow \infty} \frac{\sqrt{1+\tan ^{-1}(n)}}{n^{2}}=\frac{\sqrt{1+\tan ^{-1}(\infty)}}{\infty^{2}}$

$=\frac{\sqrt{1+\frac{\pi}{2}}}{\infty}=0$

squeeze theorem

$a n \le b n \le c{n}$

if $\lim _{n \rightarrow \infty} a n=\lim _{n \rightarrow \infty} c n=L$

$∴ \lim _{n \rightarrow \infty} b{n}=L$

$-1 \leq \sin (n) \leq 1$

$0 \le 1+\sin (n) \le 2$

$\{a n\}$

$0 \le \frac{\sqrt{1+\sin (n)}}{n^{2}} \le \frac{\sqrt{2}}{n^{2}}$

$\lim _{n \rightarrow \infty}(0)=0$

$\lim _{n \rightarrow \infty} \frac{\sqrt{2}}{n^{2}}=\frac{\sqrt{2}}{\infty^{2}}=0$

$\lim _{n \rightarrow \infty} \frac{\sqrt{1+\sin (n)}}{n^{2}}=0$

$\lim _{n \rightarrow \infty}\{a n\}=0$

convergent

Let $\{a n\}$ be a convergent sequence such that $a_{1}=\sqrt{2}$ and $a n=\sqrt{2+a_{n-1}}$ where $n \geq 2$

a) Find the first three terms of the sequence

b) Find $\lim _{n \rightarrow \infty} a n$

(a) $a_{1}=\sqrt{2}$

$a_{2}=\sqrt{2+\sqrt{2}}$

$a_{3}=\sqrt{2+\sqrt{2+\sqrt{2}}}$

$a n=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$

Since $\{a\}$ is convergent, then

$\lim _{n \rightarrow \infty} a_{n}=L=\lim _{n \rightarrow \infty} a_{n+1}=\lim _{n \rightarrow \infty} a_{n-1}=L$

$\lim _{n \rightarrow \infty} a n=\lim_{n \rightarrow \infty} \sqrt{2+a_{n-1}}=\sqrt{\lim _{n \rightarrow \infty} 2+\lim _{n \rightarrow \infty} a_{n-1}}$

$\lim _{n \rightarrow \infty} a_{n}=\sqrt{2+\lim _{n \rightarrow \infty} a_{n-1}}$

$\Rightarrow L=\sqrt{2+L} \Rightarrow L^{2}=2+L$

$+L^{2}-L-2=0 \Rightarrow(1-2)(L+1)=0$

$\Rightarrow L-2=0 \rightarrow L=2$

$L+1=0 \rightarrow L=-1$ rejected

$\lim _{n \rightarrow \infty}\{an\}=2$

Let the sequence given by the recurrence relation $2 a_{n+1}=7 \sin \left(\frac{1}{n}\right)+a_{n}+6$

where $a_{1}=2$ assuming that $a_{n}$ converges, find its limit

$\{a n\}$ converges

$\Rightarrow \lim _{n \rightarrow \infty} a_{n-1}=\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} a_{n+1}=L$

$2 a_{n+1}=7 \sin \left(\frac{1}{n}\right)+a_{n}+6$

$2 \lim _{n \rightarrow \infty} a_{n+1} =\lim _{n \rightarrow \infty}\left[7 \sin \left(\frac{1}{n}\right)+a_{n}+6\right]$

$2 L=7 \lim _{n \rightarrow \infty} \sin \left(\frac{1}{n}\right)+\lim _{n \rightarrow \infty} a_{n}+\lim _{n \rightarrow \infty} 6$

$2 L=7(0)+L+6$

$2 L=L+6 \rightarrow 2 L-L=6 \rightarrow L=6$

$\lim _{n \rightarrow \infty} a_{n}=6$

Show that the sequence $\left\{n e^{-n}\right\}$ is bounded

$a_{n}=n e^{-n}=\frac{n}{e^{n}}$

we know that $n \geq 1 \quad \& \quad e \neq 0$

we also know that $n<e^{n} \: \& \; n>\ln (n)$

$\frac{n}{e^{n}}<1$

$0<\frac{n}{e^{n}}<1$

$\left\{n e^{-n}\right\}$ is bounded below by 0 and bounded above by 1, so it is bounded