## 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

1. anonymous

Okay...think I have it.

2. anonymous

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

3. 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,

4. anonymous

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

5. anonymous

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

6. 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.

7. anonymous

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

8. anonymous

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

9. anonymous

Is that what your question wants you to do?

10. 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!!

11. 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 ;)

12. anonymous