Convergence of oscillatory/geometric series

In summary, the conversation discusses determining for which values of r the series converges and provides a solution for various cases, including when r is between 0 and 1, when r is 1 or -1, and when r is greater than or less than 1. It is concluded that the series diverges in some cases and converges absolutely in others.
  • #1
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Homework Statement


Determine for which ##r\not = 0## the series ##\displaystyle {\sum_{n=1}^\infty(2+\sin(\frac{n\pi}{3})) r^n}## converges.

Homework Equations




The Attempt at a Solution


We have to split this up by cases based on ##r##.

1) Suppose that ##0<|r|<1##. Then ##|(2+\sin(\frac{2\pi}{3}))r^n| \le |3r^n|##. But ##0<|r|<1##, so the series ##\sum_{n=1}^{\infty}3r^n## converges. By the direct comparison test, the series ##{\sum_{n=1}^\infty(2+\sin(\frac{n\pi}{3})) r^n}## converges absolutely.

2) Suppose that ##r=1##. Then ##a_n = 2+\sin(\frac{2\pi}{3})##. This sequence oscillates, so does not converge to 0. Hence the series diverges.

3) Suppose that ##r=-1##. Then let ##a_n = (-1)^n(2+\sin(\frac{n\pi}{3}))##. This sequence oscillates, so does not converge to 0. Hence the series diverges.

4) Suppose that ##r>1##. Consider the subsequence ##a_{6k-5} = (2+\frac{\sqrt{3}}{2})r^n##. For this subsequence ##\lim_{k\to\infty}a_{6k-5} = +\infty##. Hence ##\lim_{n\to\infty}a_n \not = 0##. So the series diverges.

5) Suppose that ##r<-1##. Let ##p=-r##. Then ##a_n=(-1)^n(2+\sin(\frac{2\pi}{3}))p^n##. Consider the subsequence ##a_{6k-5} = (-1)^n(2+\frac{\sqrt{3}}{2})p^n##. The limit of this sequence does not exist, because it oscillates to positive and negative infinity. Hence ##\lim_{n\to\infty}a_n \not = 0##. Hence the series diverges.
 
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  • #2
Hello mr Davis,

Any reason for this posting ? PF isn't really meant for stamp-approving homework :rolleyes: . So: where's the doubt ?
 

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