Alternating Series Test

  • Thread starter J.live
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  • #1
J.live
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Homework Statement




summation --> (-1)^n+1 (2/3)^n (I don't know how to do the symbol for sum)



The Attempt at a Solution




I) lim n-->∞ (2/3)^n = limit does not exist ? It diverges ?


P.S I am not sure if this is true. Any explanation will be a great help. Thanks.
 

Answers and Replies

  • #2
micromass
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The limit
[tex] \lim_{n\rightarrow +\infty}{(2/3)^n} [/tex]
does exist. It equals zero. So no help there.

You can in fact show that
[tex] \sum_{n=0}^n{(2/3)^n} [/tex]
converges (so the original alternating series is absolutely convergent). Can you show this? HINT: it's a geometric progression and thus the exact limit can be found...
 
  • #3
J.live
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Can you please explain how you applied n --> ∞ to (2/3)^n = (2/3) ^∞ How do i solve this?
 
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  • #4
micromass
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You have without a doubt seen that

[tex] a^n\rightarrow 0 [/tex]

if |a|< 1. Haven't you?
 
  • #5
J.live
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Yes, I am aware of the Geometric Series test. That was not my question.

My question is how did you derive to answer zero when you replaced n with ∞.

I am aware that (2/3)^n is a geometric series.I am having trouble with taking the limit of (2/3)^n.

i.e. when lim n-->∞ 1/n = 1/∞ = 0. Similarly, how do I take the limit n-->∞ in this case ?
 
  • #6
micromass
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I wasnt talking about geometric series. I was talking about the sequence [tex] (a^n)_n [/tex] with a<1. Such a sequence always has

[tex] \lim_{n\rightarrow +\infty} {a^n} = 0 [/tex]

You must have seen this.
 
  • #7
J.live
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Yes, I have seen this. Ah, makes sense. Thanks.
 
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