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Answered question

2022-03-28

Determine whether each of the following sequences with the given n-th term is convergent or divergent. Find the limit of those sequences that converge.
1) ${a}_{n}={\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\mathrm{cos}\frac{1}{n}\right)$
2) ${a}_{n}=\frac{{2}^{n+2}+5}{{3}^{n-1}}$
3) ${a}_{n}={\left(1+n\right)}^{\frac{1}{2n}}$

Answer & Explanation

Ruben Gibson

Beginner2022-03-29Added 9 answers

1) ${a}_{n}={\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\mathrm{cos}\frac{1}{n}\right)$
$\sum _{n=1}^{\mathrm{\infty }}{\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\mathrm{cos}\frac{1}{n}\right)$
Apply series divergence test
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n}\ne 0$$\sum _{n=1}^{\mathrm{\infty }}{a}_{n}⇒divergence$
$=\underset{n\to \mathrm{\infty }}{lim}{\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\mathrm{cos}\frac{1}{n}\right)$
$=\frac{1}{\sqrt{2}}\underset{n\to \mathrm{\infty }}{lim}{\mathrm{sin}}^{-1}\left(\mathrm{cos}\frac{1}{n}\right)$
Apply limit
$={\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\mathrm{cos}\frac{1}{\mathrm{\infty }}\right)$
$={\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\mathrm{cos}\left(0\right)\right)$ $\left[\mathrm{cos}\left(0\right)=1\right]$
$={\mathrm{sin}}^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$=\frac{\pi }{4}\ne 0$
Thus, ${a}_{n}$ diverges
2) ${a}_{n}=\frac{{2}^{n+2}+5}{{3}^{n-1}}$
Now, $\sum _{n=0}^{\mathrm{\infty }}\frac{{2}^{n+2}+5}{{3}^{n-1}}=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{{2}^{n+2}}{{3}^{n-1}}+\frac{5}{{3}^{n-1}}\right)$
$=\sum _{n=0}^{\mathrm{\infty }}\left(\frac{{2}^{n+2}}{{3}^{n-1}}+\sum _{n=0}^{\mathrm{\infty }}\frac{5}{{3}^{n-1}}\right)$
Apply ratio test
${a}_{n}=\frac{{2}^{n+2}}{{3}^{n-1}}$${a}_{n+1}=\frac{{2}^{n+3}}{{3}^{n}}$
${a}_{n}=\frac{5}{{3}^{n-1}}$${a}_{n+1}=\frac{5}{{3}^{n}}$
$⇒\underset{n\to \mathrm{\infty }}{lim}|\frac{{a}_{n+1}}{{a}_{n}}|=\underset{n\to \mathrm{\infty }}{lim}|\frac{{2}^{n+3}}{{3}^{n}}×\frac{{3}^{n}×{3}^{-1}}{{2}^{n}×{2}^{2}}|$
$=2×\frac{1}{3}=\frac{2}{3}<1$, converges
Similarly,
$\underset{n\to \mathrm{\infty }}{lim}|\frac{5}{{3}^{n}}×\frac{{3}^{n-1}}{5}|$
$=\frac{1}{3}<1$, converges
Thus, the sequence converges

Malia Booth

Beginner2022-03-30Added 16 answers

c) ${a}_{n}={\left(1+n\right)}^{\frac{1}{2n}}$
$\sum _{n=1}^{\mathrm{\infty }}{\left(1+n\right)}^{\frac{1}{2n}}$
Apply series divergence test
$\underset{n\to \mathrm{\infty }}{lim}{a}_{n}=\underset{n\to \mathrm{\infty }}{lim}{\left(1+n\right)}^{\frac{1}{2n}}$
Apply limit
$={\left(1+\left(\mathrm{\infty }\right)\right)}^{\frac{1}{\mathrm{\infty }}}$
$={\left(1+\mathrm{\infty }\right)}^{0}$
$=1$
$\underset{n\to \mathrm{\infty }}{lim}{\left(1+n\right)}^{\frac{1}{2n}}=\underset{n\to \mathrm{\infty }}{lim}{e}^{\frac{1}{2n}\mathrm{ln}\left(1+n\right)}=1$
As $\underset{n\to \mathrm{\infty }}{lim}{a}_{n}\ne 0$
The sequence diverges

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