I'm solving for a problem. I got this probability mass function $$ P(X=x) = \begin{pmatrix} x-1 \\ n-1 \end{pmatrix} p^n(1-p)^{x-n} \hspace{1cm}; x=n,n+1,... $$
I should find the expectation
$$ E(\frac{n-1}{x-1}) $$
I started solving it
$ E(\frac{n-1}{x-1}) = \sum_{x=n}^{\infty} \frac{n-1}{x-1} \begin{pmatrix} x-1 \\ n-1 \end{pmatrix} p^n(1-p)^{x-n} $
$ \hspace{1.6cm} = \sum_{x=n}^{\infty} \begin{pmatrix} x-2 \\ n-2 \end{pmatrix} p^n(1-p)^{x-n} $
$ \hspace{1.6cm} = p^n\sum_{x=n}^{\infty} \begin{pmatrix} x-2 \\ n-2 \end{pmatrix}(1-p)^{x-n} $
$ \hspace{1.6cm} = p^n\sum_{x=n}^{\infty} \frac{(x-2)!}{[(x-2)-(n-2)]!(n-2)!}(1-p)^{x-n} $
$ \hspace{1.6cm} = p^n\sum_{x=n}^{\infty} \frac{(x-2)!}{(x-n)!(n-2)!}(1-p)^{x-n} $
I'm stuck at this point. I don't know how to proceed. Can someone help me on how to solve or proceed further. I think Binomial Expansion might be useful but I don't know how to apply it in this case.
Taking from your las step which can be re-written as $$E=p^n\sum_{x=n}^{\infty} {x-2 \choose n-2} (1-p)^{x-n}=p^n\sum_{j=0}^{\infty} {j+n-2 \choose j} (1-p)^j=p^n[1-(1-p)]^{-(n-2)-1}=p.$$ $j=x-n$ used in above. Here we have also used the binomial series for negative integer index as: $$\sum_{J=0}^{\infty} {J+N \choose J} Z^{J}=(1-Z)^{-N-1}$$