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anonymous
 5 years ago
Use the comparison test to decide whether or not the series
summation from 1 to infinite: [n(2^n)sin n/3^n
converges
anonymous
 5 years ago
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
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...think I have it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First, let me check that your summand is \[\frac{n2^n\sin n}{3^n}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\left \frac{n2^n\sin n}{3^n} \right\le \left n \left( \frac{2}{3} \right)^n \right\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since the magnitude of sine is always less than or equal to one.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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
 5 years ago
Best ResponseYou've already chosen the best response.0Since the limiting ratio is less than 1, the series converges absolutely. Therefore, by the comparison test, your original series is convergent.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you think of a way to do this either using a pseries or a geomeetric series?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Is that what your question wants you to do?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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
 5 years ago
Best ResponseYou've already chosen the best response.0Really? Usually it's the first test taught. If I come across something, I'll let you know. Fee free to become a fan ;)
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