Need Help?

  • Notes
  • Comments & Questions

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

No comments yet

Join the conversation

Join Notatee Today!