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

In summary, the conversation discusses the convergence or divergence of the series (-2)^n/3^(n+1). It is determined that the series actually converges, despite the answer key stating otherwise. The concept of geometric series is brought up and it is mentioned that the sequence (-2)^n/3^(n+1) also converges to 0 with |r|<1.
  • #1
frasifrasi
276
0
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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  • #2
Are you sure you wrote it correct? Because the series

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

does converge...
 
  • #3
I have in the answer that it diverges...could you explain how you arrived at that?
 
  • #4
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.
 
  • #5
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)
 
  • #6
If you mean |r|<1, then yes, the sequence converges to zero.
 

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

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