Showing that $f(x)$ is a probability density function I'm having trouble showing the following $f(x)$ is a probability density function. The following information is given: 
$$f(x) = \frac{1}{\sum_{k=x}^\infty {k-1\choose x-1}(1-p)^{k-x-1}}; x = 0, 1, 2, \ldots$$
I'm having trouble in particular getting the denominator to something that I can take the integral of. I've attempted to get it to be equal to a binomial series but I eventually end up with an integral of $p^x$ which is zero. 
Thanks for any help or tips. 
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\sum_{x = 0}^{\infty}\bracks{1 \over
\bbox[1px,#ffd]{\sum_{k = x}^{\infty}{k - 1 \choose x - 1}
\pars{1 - p}^{k - x - 1}}} = 1:\ {\large ?}}$.

\begin{align}
&\mbox{Note that}\quad\bbox[10px,#ffd]{\ds{%
\sum_{k = x}^{\infty}{k - 1 \choose x - 1}\pars{1 - p}^{k - x - 1}}} =
\sum_{k = 0}^{\infty}{k + x - 1 \choose x - 1}\pars{1 - p}^{k - 1}
\\[5mm] = &\
{1 \over 1 - p}\sum_{k = 0}^{\infty}{k + x - 1 \choose k}\pars{1 - p}^{k} =
{1 \over 1 - p}\sum_{k = 0}^{\infty}{-x \choose k}\pars{-1}^{k}\pars{1 - p}^{k} \\[5mm] = &\
{1 \over 1 - p}\sum_{k = 0}^{\infty}{-x \choose k}\pars{-1 + p}^{k} =
{1 \over 1 - p}\bracks{1 + \pars{-1 + p}}^{\,-x} = \bbox[10px,#ffd]{\ds{{1 \over \pars{1 - p}p^{x}}}}
\end{align}

Then,
\begin{align}
\sum_{x = 0}^{\infty}\bracks{1 \over
\bbox[1px,#ffd]{\sum_{k = x}^{\infty}{k - 1 \choose x - 1}
\pars{1 - p}^{k - x - 1}}} & =
\sum_{x = 0}^{\infty}{1 \over
\bbox[#ffd]{\ds{1/\bracks{\pars{1 - p}p^{x}}}}}
\\[5mm] & =
\pars{1 - p}\sum_{x = 0}^{\infty}p^{x} = \pars{1 - p}\,{1 \over 1 - p} = \bbx{1}
\end{align}
A: After a bit of a struggle, I figured out how to get the function to be $(1 - p)p^x$. 
By expanding the summation in the denominator and simplifying, I got: 
$(1-p)^{-1} + x(1-p)^0 + \frac{(x+1)x}{2!}(1-p) + \dots$
Then, I realized this was equal to: 
$\sum_{n=0}^{\infty}(1-p)^{n-1}{n+(x-1)\choose n}$
This is then equal to:  
$(1-p)^{-1}\sum_{n=0}^{\infty}(1-p)^{n}{n+(x-1)\choose n}$
The summation is then equal to the general binomial series and f(x) becomes: 
$\frac{1}{\frac{1}{(1-p)(1-(1-p)^{x-1+1}}} \text{ or } (1-p)(1-(1-p))^{x-1+1}$ 
Which simplifies to what we wanted and it was easy enough to prove from there that the summation of f(x) from 0 to infinity was 1. 
A: $\color{blue}{\text{I apologize for the incomplete solution, so I would appreciate anyone that can add in to help out}}$
What we want to show is $$\sum_{x=0}^\infty \left(\frac{1}{\sum_{k=x}^{\infty} {k-1\choose x-1}(1-p)^{k-x-1}}\right) = 1$$
The first step could be to reduce the fraction into something we can work with.
It appears that $$\frac{1}{\sum_{k=x}^{\infty} {k-1\choose x-1}(1-p)^{k-x-1}} = (1-p)p^x$$
WolframAlpha confirms this, but I would appreciate if anyone else can explain why, as I don't know how to show this by hand
Once that is shown, then it isn't too hard to show that $(1-p)\sum_{x=0}^{\infty}p^x = 1 $ since we are now dealing with a geometric series.
