What is the sum of this infinite series? Which one is it, Taylors? Binomial? I am trying to figure which formula to use for this one.
$$\displaystyle\sum\limits_{x=-1}^{-\infty} -x(1-y)p(1-p)^{-x}+\sum\limits_{x=1}^\infty xyp(1-p)^x$$
where $0<y<1$, and $0<p<1$.
My work:
$$=\displaystyle\sum\limits_{x=1}^{\infty} -x(1-y)p(1-p)^{x}+\sum\limits_{x=1}^\infty xyp(1-p)^x$$
next step,
$$=\displaystyle\sum\limits_{x=1}^{\infty} p(1-p)^{x}(2xy-x)$$
This doesn't look like anything.  Does anyone recognize this?
 A: We have, provided $|a|<1$, that $$\sum_{n=0}^\infty na^n=\frac{a}{(a-1)^2}$$
In your problem, we may rewrite $$\sum_{x=1}^\infty p(1-p)^x(2xy-x)=(2y-1)p\sum_{x=1}^\infty x(1-p)^x$$
Applying that first identity, we find the sum to be $$\frac{(2y-1)p(1-p)}{p^2}=\frac{(2y-1)(1-p)}{p}$$
Edit: Deriving the formula, per request.
We start with the geometric series $$\sum_{n=0}^\infty a^n=\frac{1}{1-a}$$
We take $\frac{d}{da}$ of both sides to get $$\sum_{n=0}^\infty na^{n-1}=\frac{1}{(1-a)^2}$$
We multiply both sides by $a$ to get $$\sum_{n=0}^\infty na^{n}=\frac{a}{(1-a)^2}$$
A: It works like @vadium123 mentioned, I just wanna give you the idea how one derives that result.
The first idea is that 
\[ \sum_{n=0}^\infty n \cdot x^n  \]
looks like the goemetric series (at least a bit). In fact if we take the derivative the geometric series in respect to $x$ we get
\[ \sum_{n=1}^\infty n\cdot x^{n-1} \] 
So when we take 
\[ x \cdot \sum_{n=1}^\infty n \cdot x^{n-1} = \sum_{n=1}^\infty n\cdot x^n\]
Furthermore we know that 
\[ \sum_{n=0}^\infty x^n = \frac{1}{1-x} \]
When we take the derivative on both sides we have
\[ \frac{1}{(1-x)^2}\]
As we need to multiply with $x$ our final sum will be 
\[ \frac{x}{(1-x)^2}\]
A: First, this sum is an instance of the binomial theorem and $\binom{k}{k-1}=(-1)^{k-1}\binom{-2}{k-1}$
$$
\begin{align}
\sum_{k=1}^\infty kx^k
&=\sum_{k=1}^\infty\binom{k}{1}x^k\\
&=\sum_{k=1}^\infty\binom{k}{k-1}x^k\\
&=\sum_{k=1}^\infty(-1)^{k-1}\binom{-2}{k-1}x^k\\
&=\sum_{k=0}^\infty(-1)^k\binom{-2}{k}x^{k+1}\\
&=x\sum_{k=0}^\infty\binom{-2}{k}(-x)^k\\[4pt]
&=x(1-x)^{-2}\\[9pt]
&=\frac{x}{(1-x)^2}\tag{1}
\end{align}
$$
Now to apply $(1)$, we simply reindex to positive indices ($k\mapsto-k$). I believe that this was done incorrectly in the question. Here $x=1-p$.
$$
\begin{align}
&\sum\limits_{k=-1}^{-\infty} -k(1-y)p(1-p)^{-k}+\sum\limits_{k=1}^\infty kyp(1-p)^k\\
&=\sum\limits_{k=1}^{\infty} k(1-y)p(1-p)^k+\sum\limits_{k=1}^\infty kyp(1-p)^k\\
&=\sum\limits_{k=1}^{\infty} kp(1-p)^k\\
&=\frac{p(1-p)}{(1-(1-p))^2}\\
&=\frac{1-p}{p}\tag{2}
\end{align}
$$
