How to evaluate $1 - \frac{\binom{n^2}{1}}{\binom{n+1}{1}} + \frac{\binom{n^2}{2}}{\binom{n+2}{2}} - \frac{\binom{n^2}{3}}{\binom{n+3}{3}} + ..$ How to evaluate $1 - \frac{\binom{n^2}{1}}{\binom{n+1}{1}} + \frac{\binom{n^2}{2}}{\binom{n+2}{2} } - \frac{\binom{n^2}{3}}{\binom{n+3}{3}} + \frac{\binom{n^2}{4}}{\binom{n+4}{4}} - ......$ 
I really have no idea how to proceed in this question. Expanding it by the formula isn't helping as much as I can see. And the options are extremely sofisticated as well like 1/n, 1/(n+1), 1. How to proceed?


A variation. We obtain
  \begin{align*}
\color{blue}{\sum_{j=0}^{n^2}}&\color{blue}{(-1)^j\binom{n^2}{j}\binom{n+j}{j}^{-1}}\\
&=\sum_{j=0}^{n^2}\binom{n^2}{j}\binom{-n-1}{j}^{-1}\tag{1}\\
&=\sum_{j=0}^{n^2}\binom{n^2}{j}(-n)\int_{0}^1t^j(1-t)^{-n-1-j}\,dt\tag{2}\\
&=(-n)\int_{0}^1(1-t)^{-n-1}\sum_{j=0}^{n^2}\binom{n^2}{j}\left(\frac{t}{1-t}\right)^j\,dt\\
&=(-n)\int_{0}^1(1-t)^{-n-1}\left(1+\frac{t}{1-t}\right)^{n^2}\,dt\\
&=(-n)\int_{0}^1(1-t)^{-n^2-n-1}\,dt\\
&=\frac{-n}{-n^2-n}\\
&\color{blue}{=\frac{1}{n+1}}
\end{align*}

Comment:


*

*In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

*In (2) we apply the identity $\binom{n}{r}^{-1}=(n+1)\int_{0}^1t^r(1-t)^{n-r}\,dt$.
