Convergence of the Alternating Series: Investigating (2/3)^n Summation

[J.live]

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.

 

[micromass]

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

You can in fact show that
\sum_{n=0}^n{(2/3)^n}
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...

 

[J.live]

Can you please explain how you applied n --> ∞ to (2/3)^n = (2/3) ^∞ How do i solve this?

 

[micromass]

You have without a doubt seen that

a^n\rightarrow 0

if |a|< 1. Haven't you?

 

[J.live]

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 ?

 

[micromass]

I wasnt talking about geometric series. I was talking about the sequence (a^n)_n with a<1. Such a sequence always has

\lim_{n\rightarrow +\infty} {a^n} = 0

You must have seen this.

 

[J.live]

Yes, I have seen this. Ah, makes sense. Thanks.