New Notes Blog: `a_n = nsin(1/n)` Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

Wednesday, 11 March 2015

$\displaystyle{a}_{{n}}={n}{\sin{{\left(\frac{{1}}{{n}}\right)}}}$ Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

$\displaystyle{a}_{{n}}={n}{\sin{{\left(\frac{{1}}{{n}}\right)}}}$

Apply n'th term test for divergence, which states that,

If $\displaystyle\lim_{{{n}\to\infty}}{a}_{{n}}\ne{0}$ , then $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}$ diverges

$\displaystyle\lim_{{{n}\to\infty}}{n}{\sin{{\left(\frac{{1}}{{n}}\right)}}}=\lim_{{{n}\to\infty}}\frac{{\sin{{\left(\frac{{1}}{{n}}\right)}}}}{{\frac{{1}}{{n}}}}$

Apply L'Hospital's rule,

Test L'Hospital condition: $\displaystyle\frac{{0}}{{0}}$

$\displaystyle=\lim_{{{n}\to\infty}}\frac{{\frac{{d}}{{{d}{n}}}{\sin{{\left(\frac{{1}}{{n}}\right)}}}}}{{\frac{{d}}{{{d}{n}}}\frac{{1}}{{n}}}}$

$\displaystyle=\lim_{{{n}\to\infty}}\frac{{{\cos{{\left(\frac{{1}}{{n}}\right)}}}{\left(-{{n}}^{{-{2}}}\right)}}}{{-{{n}}^{{-{2}}}}}$

$\displaystyle=\lim_{{{n}\to\infty}}{\cos{{\left(\frac{{1}}{{n}}\right)}}}$

$\displaystyle\lim_{{{n}\to\infty}}\frac{{1}}{{n}}={0}$

$\displaystyle\lim_{{{u}\to{0}}}{\cos{{\left({u}\right)}}}={1}$

By the limit chain rule,

$\displaystyle={1}\ne{0}$

So, by the divergence test criteria series diverges.

$\displaystyle{a}_{{n}}={n}{\sin{{\left(\frac{{1}}{{n}}\right)}}}$

Apply n'th term test for divergence, which states that,

If $\displaystyle\lim_{{{n}\to\infty}}{a}_{{n}}\ne{0}$ , then $\displaystyle{\sum_{{{n}={1}}}^{\infty}}{a}_{{n}}$ diverges

$\displaystyle\lim_{{{n}\to\infty}}{n}{\sin{{\left(\frac{{1}}{{n}}\right)}}}=\lim_{{{n}\to\infty}}\frac{{\sin{{\left(\frac{{1}}{{n}}\right)}}}}{{\frac{{1}}{{n}}}}$

Apply L'Hospital's rule,

Test L'Hospital condition: $\displaystyle\frac{{0}}{{0}}$

$\displaystyle=\lim_{{{n}\to\infty}}\frac{{\frac{{d}}{{{d}{n}}}{\sin{{\left(\frac{{1}}{{n}}\right)}}}}}{{\frac{{d}}{{{d}{n}}}\frac{{1}}{{n}}}}$

$\displaystyle=\lim_{{{n}\to\infty}}\frac{{{\cos{{\left(\frac{{1}}{{n}}\right)}}}{\left(-{{n}}^{{-{2}}}\right)}}}{{-{{n}}^{{-{2}}}}}$

$\displaystyle=\lim_{{{n}\to\infty}}{\cos{{\left(\frac{{1}}{{n}}\right)}}}$

$\displaystyle\lim_{{{n}\to\infty}}\frac{{1}}{{n}}={0}$

$\displaystyle\lim_{{{u}\to{0}}}{\cos{{\left({u}\right)}}}={1}$

By the limit chain rule,

$\displaystyle={1}\ne{0}$

So, by the divergence test criteria series diverges.

New Notes Blog

Wednesday, 11 March 2015

$\displaystyle{a}_{{n}}={n}{\sin{{\left(\frac{{1}}{{n}}\right)}}}$ Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Wednesday, 11 March 2015

Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{a}_{{n}}={n}{\sin{{\left(\frac{{1}}{{n}}\right)}}}$ Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?