<s>[INST] Trouble with the Binomial sum I guess the following sum is equal to $1$ since you can compute the baby cases for $t=1,2$ straightforward to get the desired result. So, in general, we should have the following sum to be $1$, I think. But, I am stuck in the last equality. Any comments or advice would be appreciated.
\begin{align*}
    &\frac{t}{n}+t\left(\frac{\binom{n-t}{1}}{2\binom{n}{2}}+\frac{\binom{n-t}{2}}{3\binom{n}{3}}+\frac{\binom{n-t}{3}}{4\binom{n}{4}}+\cdots+\frac{\binom{n-t}{n-t}}{(n-t+1)\binom{n}{n-t+1}}\right)\\\\
    =~~&\frac{t}{n}+t\left(\frac{\binom{n-t}{1}}{n\binom{n-1}{1}}+\frac{\binom{n-t}{2}}{n\binom{n-1}{2}}+\frac{\binom{n-t}{3}}{n\binom{n-1}{3}}+\cdots+\frac{\binom{n-t}{n-t}}{n\binom{n-1}{n-t}}\right)\\\\
    =~~&\frac{t}{n}+\frac{t}{n}\left(\frac{\binom{n-t}{1}}{\binom{n-1}{1}}+\frac{\binom{n-t}{2}}{\binom{n-1}{2}}+\frac{\binom{n-t}{3}}{\binom{n-1}{3}}+\cdots+\frac{\binom{n-t}{n-t}}{\binom{n-1}{n-t}}\right)
\end{align*}
where the first equality is used $k\binom{n}{k}=n\binom{n-1}{k-1}$ and $t\in\{1,2,...,n\}.$
 [/INST] Let us find the summation
$$S_{m,n}=\sum_{k=1}^{m} \frac{m \choose k}{(k+1){n\choose k+1}}$$
Use
${N\choose K}^{-1}=(N+1)\int_{0}^{1} x^K(1-x)^{N-K} dx$
$$S_{m,n}=(n+1)\sum_{k=1}^m\frac{m\choose k}{k+1} x^{k+1}(1-x)^{n-k-1}dx$$
Let $\frac{x}{1-x}=z$,
$$S_{m,n}=(n+1)\int_0^1 (1-x)^n \sum_{k=1}^{m} {m \choose k} \frac{z^{k+1}}{k+1}dx$$
By integration of $\sum_{k=1}^{m} {m\choose k} t^k=(1+t)^m-1$ from $t=0$ to $t=z$, we can re-write
$$S_{m,n}=(n+1)\int_{0}^{1}(1-x)^n \left(\frac{(1-x)^{-m-1}-1
}{m+1}-\frac{x}{1-x}\right)dx$$
$$S_{m,n}=(n+1)\int_{0}^1 \left[\frac{(1-x)^{n-m-1}}{m+1}-\frac{(1-x)^n}{m+1}-x(1-x)^{n-1}\right] dx=\frac{1}{n-m}$$
$$S_{m,n}=(n+1)\left[\frac{1}{(n-m)(m+1)}-\frac{1}{(n+1)(m+1)}\right]-\frac{1}{n}=\frac{m}{(n-m)n}.$$
Using this and $m=n-t$ the required expression is
$$\frac{t}{n}+tS_{m,n}=1$$
</s>