I need help solving the following summation:
$$2^{x-1}\sum_{y=x}^{\infty} \frac y{2^y} $$
for $y\ge x\ge 1$
I tried writing out terms to solve but that did not go anywhere. I'm not sure what else I should try or if this fits a certain summation type (i.e. geometric).
We obtain \begin{align*} \color{blue}{2^{x-1}\sum_{y=x}^\infty\frac{y}{2^y}}&=2^{x-2}\sum_{y=x}^\infty y\left(\frac{1}{2}\right)^{y-1}\tag{1}\\ &=2^{x-2}\sum_{y=x}^\infty\left.\left(\frac{d}{dt}t^y\right)\right|_{t=1/2}\tag{2}\\ &=2^{x-2}\left.\left(\frac{d}{dt}\sum_{y=x}^\infty t^y\right)\right|_{t=1/2}\tag{3}\\ &=2^{x-2}\left.\left[\frac{d}{dt}\left(\frac{t^x}{1-t}\right)\right]\right|_{t=1/2}\tag{4}\\ &=2^{x-2}\cdot\left.\frac{t^{x-1}(-tx+t+x)}{(1-t)^2}\right|_{t=1/2}\tag{5}\\ &\color{blue}{=x+1}\tag{6} \end{align*}
Comment:
In (1) we do a rearrangement and prepare for introducing differentiation.
In (2) we write the expression with the help of the differential operator $\frac{d}{dt}$ evaluated at $t=\frac{1}{2}$.
In (3) we exchange series and differential operator which is valid for $|t|<1$.
In (4) we apply the geometric series expansion.
In (5) we do the differentiation.
In (6) we evaluate at $t=\frac{1}{2}$ and simplify.
It is well-known that $$\sum_{n=1}^\infty nr^n = \frac{r}{(1-r)^2}.\tag1$$ There are various ways to prove this, including the use of a differential as in this answer. See How can I evaluate $\sum_{n=0}^\infty (n+1)x^n$. This is a variation on the even better-known sum of a geometric series, $$\sum_{n=0}^\infty r^n = \frac1{1-r}.$$ Notice that if we start this last sum from $n=1$ instead of $n=0,$ we get $$\sum_{n=1}^\infty r^n = r\sum_{n=0}^\infty r^n =\frac{r}{1-r}.\tag2$$
In the question, the index of the summation starts from $x,$ not from $1.$ We can address this in either of two ways: generalize $(1)$ so that the index can start from any integer, or change the indexing in the summation in the question. Let's change the indexing by making the substitution $y = u + x - 1.$ Then $u=1$ when $y=x,$ and
\begin{align} 2^{x-1}\sum_{y=x}^{\infty} \frac y{2^y} &= 2^{x-1}\sum_{u=1}^{\infty} \frac{u+x-1}{2^{u+x-1}} \\ &= 2^{x-1}\sum_{u=1}^{\infty} \left(\frac1{2^{x-1}}\frac{u+x-1}{2^u}\right)\\ &= \sum_{u=1}^{\infty} \frac{u+x-1}{2^u}\\ &= \sum_{u=1}^{\infty} \frac{u}{2^u} + (x-1)\sum_{u=1}^{\infty} \frac1{2^u}\\ &= \frac{\left(\frac12\right)}{\left(1-\frac12\right)^2} + (x-1)\sum_{u=1}^{\infty} \frac1{2^u} & \text{application of $(1)$} \\ &= 2 + (x-1)\sum_{u=1}^{\infty} \frac1{2^u} \\ &= 2 + (x-1)\frac{\left(\frac12\right)}{1-\frac12} & \text{application of $(2)$} \\ &= 2 + (x-1) \\ &= x + 1. \end{align}
If expanded:
$$2^{x-1}\sum_{y=x}^{\infty} \frac y{2^y}=2^{x-1}\left(\frac{x}{2^x}+\frac{x+1}{2^{x+1}}+\frac{x+2}{2^{x+2}}+\cdots\right)=\frac{x}{2}+\frac{x+1}{2^2}+\frac{x+2}{2^3}+\cdots=$$ $$x\left(\underbrace{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots}_{=1}\right)+\left(\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\cdots\right)=$$ $$x+\left([\underbrace{\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\cdots}_{=\frac{1}{2}}]+\underbrace{[\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\cdots}_{\frac{1}{2^2}}]+\underbrace{[\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\cdots}_\frac{1}{2^3}]+\cdots\right)=$$ $$x+\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots\right)=x+1.$$
