Calculate:- $$\sum_{r=1}^{2023} \frac{(-1)^{r-1}r}{2024 \choose r}$$ And generalise the result if possible.
I've tried to reduce this to a telescopic sum but could not do it.
I've also made a recurrence relation:-
If S$_{n}$ = $\sum_{r=1}^{n-1} \frac{(-1)^{r-1}r}{n \choose r}$, then S$_n$ = $\frac{n+1}{2n}$$S_{n-1}$ + $\frac{{(-1)}^n}{n}$, but could not go further.
$$\sum_{i=1}^{2023} \frac{(-1)^{r-1}r}{2024 \choose r}= \frac{(-1)^{r-1}r}{2024 \choose r}\sum_{i=1}^{2023} 1=\frac{(-1)^{r-1}r \,2023}{2024 \choose r}.$$