Question

`sum_(k=0)^oo(-1)^k ((k+1)/4^(k+1)) =` ?

Answer 1

See below.

Explanation:

`1/(x+1)^2= -d/(dx)(1/(x+1))` but

`1/(x+1)=1/(2+x-1)=1/2(1/(1+(x-1)/2)) = 1/2sum_(k=0)^oo(-1)^k((x-1)/2)^k` for `abs((x-1)/2)<1`

but

`-d/(dx)(1/2sum_(k=0)^oo(-1)^k((x-1)/2)^k)=-1/2^2sum_(k=1)^oo(-1)^k k/2^(k-1)(x-1)^(k-1) = sum_(k=0)^oo(-1)^k ((k+1)/2^(k+2))(x-1)^k`

Making `x = 1/2` we have

`sum_(k=0)^oo(-1)^k ((k+1)/4^(k+1)) = 1/(1/2+1)^2 = (2/3)^2`