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

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Answer 1 · 2017-03-03T16:03:36

See below.

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

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

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