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- anonymous

Use the comparison test to decide whether or not the series
summation from 1 to infinite: [n(2^n)|sin n|/3^n
converges

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- anonymous

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- anonymous

Okay...think I have it.

- anonymous

First, let me check that your summand is \[\frac{n2^n|\sin n|}{3^n}\]

- anonymous

The comparison test says that if you have a series with terms a_n, and for sufficiently large n, you have b_n such that |a_n|<=|b_n|, and where the series for b_n is absolutely convergent, then you original series is absolutely convergent, and therefore, the original series is convergent. Now,

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- anonymous

\[\left| \frac{n2^n|\sin n|}{3^n} \right|\le \left| n \left( \frac{2}{3} \right)^n \right|\]

- anonymous

since the magnitude of sine is always less than or equal to one.

- anonymous

The series given by \[\sum_{n=0}^{\infty} n \left( \frac{2}{3} \right)^n\] is absolutely convergent, since, by the ratio test,\[\frac{\left| (n+1)\left( \frac{2}{3} \right)^{n+1} \right|}{\left| (n)\left( \frac{2}{3} \right)^{n} \right|}=\frac{2}{3}\frac{n+1}{n}=\frac{2}{3}\frac{1+\frac{1}{n}}{1}\]which goes to 2/3 as n goes to infinity.

- anonymous

Since the limiting ratio is less than 1, the series converges absolutely. Therefore, by the comparison test, your original series is convergent.

- anonymous

can you think of a way to do this either using a p-series or a geomeetric series?

- anonymous

Is that what your question wants you to do?

- anonymous

i just haven't learnt the ratio test yet, so im wondering if there are other ways to do it. this should work though. thanks for your help!!

- anonymous

Really? Usually it's the first test taught.
If I come across something, I'll let you know.
Fee free to become a fan ;)

- anonymous

already done! thanks

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