anonymous
  • 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
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
Okay...think I have it.
anonymous
  • anonymous
First, let me check that your summand is \[\frac{n2^n|\sin n|}{3^n}\]
anonymous
  • 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
  • anonymous
\[\left| \frac{n2^n|\sin n|}{3^n} \right|\le \left| n \left( \frac{2}{3} \right)^n \right|\]
anonymous
  • anonymous
since the magnitude of sine is always less than or equal to one.
anonymous
  • 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
  • anonymous
Since the limiting ratio is less than 1, the series converges absolutely. Therefore, by the comparison test, your original series is convergent.
anonymous
  • anonymous
can you think of a way to do this either using a p-series or a geomeetric series?
anonymous
  • anonymous
Is that what your question wants you to do?
anonymous
  • 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
  • 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
  • anonymous
already done! thanks

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