calculus
posted by bill .
The sum from 0 to infinity of (1)^n(3/3^n) is convergent or divergent? If convergent, what is the sum?
I got that it's convergent and the sum is 9/2, but that's wrong.

It is not clear without sufficient parentheses what the expression really is.
I assume it to be:
Sum((1)^n * (3/3^n)) for n=0 > ∞
This is an alternating geometric series.
(9/2) is the correct sum for the geometric series (nonalternating).
Write out the first few terms of the series:
3  3/3 + 3/9  3/27 + 3/81  ...
which can be regrouped into two geometric series:
3(1+1/9+1/81+...)  (1+1/9+1/81+...)
=3(9/8)  (9/8)
=2(9/8)
=9/4
Note that if the minus sign becomes a plus sign, we get the geometric sum of 9/2. 
Oh, right . I forgot all about the 1. Thanks.

You're welcome!
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