# Alternating Series Test

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.

## Answers and Replies

Staff Emeritus
Homework Helper
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?

Last edited:
Staff Emeritus
Homework Helper
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 ?

Staff Emeritus
I wasnt talking about geometric series. I was talking about the sequence $$(a^n)_n$$ with a<1. Such a sequence always has
$$\lim_{n\rightarrow +\infty} {a^n} = 0$$