calculus
posted by sarah .
find the sum of the series:
1. the sum from n=1 to infinity of ((1)^n*(.2)^n)/n
I simplified this to: (.2)^n/n
I know this is alternating, but how do I know what the sum is?
2. the sum from n=0 to infinity of 1/2^n
Is this geometric with n^(2)? and if so, do you use the formula?
3. the sum from n=1 to infinity of n/(2^(n1))
THANK YOU SO MUCH

calculus 
Count Iblis
1) Log[1 + x] =
sum from n = 1 to infinity of
(1)^(n+1) x^n/n
2) 1/[1+x] = sum from n=0 to infinity of x^n
x = 1/2, there is no n^(2) in here.
3)
Differentiate both sides of the equation:
1/[1+x] = sum from n=0 to infinity of x^n
with respect to x.
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