Simplifying ratio test with factorials and exponentials

quickclick330 · Nov 13, 2007

Don't understand how they simplified this...(ratio test)

I'm studying for my exam and I was looking at this example:

Typed worked solution of a ratio test: limit as n goes to infinity of |a_(n+1)/a_n| equals the limit of |(-3)^(n+2) times 2^(3n) over 2^(3(n+1)) times (-3)^(n+1)|, simplifying to 3/8 < 1, so the series is absolutely convergent.

I'm not really sure how they get from here :

Limit expression: = lim as n approaches infinity of the absolute value of [(-3)^(n+2) / 2^(3(n+1))] times [2^(3n) / (-3)^(n+1)].

to here:

$Limit as n approaches infinity of the absolute value of (-3 times 2 to the power 3n) divided by (2 to the power 3n times 2 to the power 3), written as a plain fraction inside absolute value bars.$

If someone could explain how they simplified this to me that would be fantastic...i'm really trying to understand series but sometimes the algebra stuff holds me up :-(.

SNOOTCHIEBOOCHEE · Nov 13, 2007

You are basically just dividing like terms, so you subtract exponents

(-3)^n+2 / (-3)^n+1 = (-3)^ ((n+2)-(n+1)) = -3

2^3n / 2^(3n+3) = 2^((3n) -(3n+3)) = 2^-3

hope that helps

quickclick330 · Nov 13, 2007

yes that helps...thank you :-)

quickclick330 · Nov 13, 2007

actually that helped a lot... thank you thank you!

Simplifying ratio test with factorials and exponentials

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