Does the Series Sum of (-2)^n / 3^(n+1) from 0 to Infinity Diverge?

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The discussion revolves around the convergence or divergence of the series summation of the expression (-2)^(n) / 3^(n+1) from 0 to infinity, which falls under the topic of infinite series and geometric series.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the nature of the series, questioning whether it diverges due to the alternating signs of the terms. Some participants suggest that the series may actually converge, referencing its geometric series form.

Discussion Status

The discussion includes differing opinions on the convergence of the series, with some participants asserting that it converges while others maintain that it diverges. There is an exploration of how to analyze the sequence derived from the series and its convergence to zero.

Contextual Notes

There is mention of an answer key that claims divergence, which some participants question. The discussion also touches on the conditions for convergence of sequences and series, particularly in relation to the ratio of terms.

frasifrasi
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can anyone explain why the summation from 0 to infinity of

(-2)^(n)/3^(n+1) diverges?

- Is it simply because the terms bounce between - and +?
 
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Are you sure you wrote it correct? Because the series

[tex]\sum_{n=0}^{\infty} \frac{(-2)^n}{3^{n+1}}[/tex]

does converge...
 
I have in the answer that it diverges...could you explain how you arrived at that?
 
frasifrasi said:
I have in the answer that it diverges...could you explain how you arrived at that?

It's (1/3)*(-2/3)^n. It's a geometric series. You can even sum it.
 
I see the light! I guess the answer key was wrong. But hey,

What if had the SEQUENCE (-2)^(n)/3^(n+1) , how could I show that it converges to 0?

could I also "simplify" it to (1/3)*(-2/3)^n and say that since r > -1, it converges to 0?

(by the fact that for a sequence r^n , the sequence converges for -1 < r <= 1)
 
If you mean |r|<1, then yes, the sequence converges to zero.
 

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