Question

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

Answer 1

Okay...think I have it.

Answer 2

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

Answer 3

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,

Answer 4

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

Answer 5

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

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

Is that what your question wants you to do?

Answer 10

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!!

Answer 11

Really? Usually it's the first test taught.

If I come across something, I'll let you know.

Fee free to become a fan ;)

Answer 12

already done! thanks