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: Below is a quite brutal approach : whenever I see inverses of binomial coefficients, I try to use the following relation between the Gamma and Beta functions (I can provide a link if needed) :

$$\mbox{With  } 0 < m \le n,\quad \quad \ \frac{1}{\binom{m+n}{m}} = \frac{mn}{m+n} \cdot \displaystyle{\int_0^1} t^{m-1}(1-t)^{n-1} dt$$

Your sum becomes : 
$$S_n = 1 + \sum \limits_{k=1}^{n^2} (-1)^k \binom{n^2}{k} \frac{kn}{k+n} \displaystyle{\int_0^1} t^{k-1}(1-t)^{n-1} dt$$
$$S_n = 1 + n \displaystyle{\int_0^1} (1-t)^{n-1} \cdot \Big( \sum \limits_{k=1}^{n^2} \frac{(-1)^k k}{k+n} \binom{n^2}{k} t^{k-1} \Big) dt$$
Short interlude : we can simplify this sum. Denote $f : t \mapsto t^{n+1} \sum \limits_{k=1}^{n^2} \frac{(-1)^k k}{k+n} \binom{n^2}{k} t^{k-1}$. Then $f'(t) = \sum \limits_{k=1}^{n^2} (-1)^k k \binom{n^2}{k} t^{k+n-1} = t^n \frac{d}{dt} \big( \sum \limits_{k=0}^{n^2} \binom{n^2}{k} (-t)^k \Big) = t^n \frac{d}{dt} \Big( (1-t)^{n^2} \Big) = -n^2 t^n (1-t)^{n^2-1}$.
Hence $$S_n = 1 + n \displaystyle{\int_0^1} (1-t)^{n-1} \cdot \frac{1}{t^{n+1}} \displaystyle{\int_0^{t}}-n^2 s^n (1-s)^{n^2-1} ds dt$$
and by exchanging the integrals : 
$$S_n = 1 - n^3 \displaystyle{\int_0^1} s^n (1-s)^{n^2-1} \cdot \displaystyle{\int_s^1} \frac{(1-t)^{n-1}}{t^{n+1}} dtds$$
$$S_n = 1 - n^3 \displaystyle{\int_0^1} s^n (1-s)^{n^2-1} \cdot \frac{(1-s)^n}{n s^n}ds$$
$$S_n = 1 -n^2 \displaystyle{\int_0^1} (1-s)^{n^2+n-1} = 1 - \frac{n^2}{n^2+n}$$
There might (must) be a more elegant way, but with this we can conclude that $S_n = \frac{1}{n+1}$.
A: 
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 $\binom{n}{r}^{-1}=(n+1)\int_{0}^1t^r(1-t)^{n-r}\,dt$.
