# calculus

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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^(n-1))

THANK YOU SO MUCH

• calculus -

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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