Question

why does the summation of (i/2^i) from 0 to infinity sum to 2? Is this a special summation? Am trying to figure out the intuition here. Thanks!

Answer 1

\[\sum_{i=0}^{\infty}\frac{i}{2^i}\]

Answer 2

consider that we know $$S=1+x+x^2+\dots\\xS=x+x^2+x^3+\dots\\S-xS=1\\(1-x)S=1\\S=\frac1{1-x}$$ when \(|x|<1\) so that \(S\) converges. in other words, $$\sum_{k=0}^\infty x^k=\frac1{1-x}$$now, differentiate both sides (for the series, take the derivative term by term): $$\sum_{k=0}^\infty kx^{k-1}=\frac1{(1-x)^2}\\\sum_{k=0}^\infty kx^k=\frac{x}{(1-x)^2}$$by multiplying both sides by \(x\)

Answer 3

now plug in \(x=1/2\): $$\sum_{k=0}^\infty k\left(\frac12\right)^k=\frac{1/2}{(1-1/2)^2}\\\sum_{k=0}^\infty k\cdot\frac{1}{2^k}=\frac{1/2}{1/4}\\\implies\sum_{k=0}^\infty\frac{k}{2^k}=2$$

Answer 4

alternatively, we can break it down like: $$\begin{align*}\sum_{k=0}^\infty\frac{i}{2^i}&=0+1\cdot\frac12+2\cdot\frac14+3\cdot\frac18+\dots\\&=\left(\frac12+\frac14+\frac18+\dots\right)+\left(\frac14+\frac18+\dots\right)+\left(\frac18+\dots\right)+\dots\\&=\left(1+\frac12+\frac14+\dots\right)\left(\frac12+\frac14+\dots\right)\\&=\frac1{1-1/2}\cdot\frac12\cdot\frac1{1-1/2}\\&=2\cdot\frac12\cdot2\\&=2\end{align*}$$

Answer 5

awesome, thanks that helps a lot, especially the first explanation.