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August 22, 2014

August 22, 2014

Posted by **Lucy** on Monday, October 13, 2008 at 11:55am.

a) 1+3(1/4)+9(1/4)^2+27(1/4)^3...

b) 1+3(1/5)+9(1/5)^2+27(1/5)^3...

c) 1+3(1/7)+9(1/7)^2+27(1/7)^3...

d) 1+3(1/2)+9(1/2)^2+27(1/2)^3...

How do you determine if a series in convergent or divergent???

The book that I have is about as clear as mud and I do not understand...

Thanks for your help!!

- Pre-calculus -
**Reiny**, Monday, October 13, 2008 at 12:23pmThis would be a lengthy reply, so see if you can make sense out of this

http://faculty.msmary.edu/heinold/m248spring2007/m248spring2007_convergence_tests.pdf

I will take a look at the first question

1+3(1/4)+9(1/4)^2+27(1/4)^3...

= 1 + 3^1 / 4^1 + 3^2 / 4^2 + ..

= 1 + summation of (3/4)^n as n goes from 1 to infinitity

Now compare this result with what it says about the summation of a geometric series at the top of the link I gave you.

good luck.

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