Calculate:- $$\sum_{i=1}^{2023} \frac{(-1)^{r-1}r}{2024 \choose r}$$ 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$_{n}$ = $\sum_{i=1}^{n-1} \frac{(-1)^{r-1}r}{n \choose r}$, then S$_n$ = $\frac{n+1}{2n}$$S_{n-1}$ + $\frac{{(-1)}^n}{n}$, but could not go further.

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.

Calculate:- $$\sum_{i=1}^{2023} \frac{(-1)^{r-1}r}{2024 \choose r}$$ And generalise the result if possible.

I've tried to reduce this to a telescopic sum but could not do it.

Calculate:- $$\sum_{i=1}^{2023} \frac{(-1)^{r-1}r}{2024 \choose r}$$ 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$_{n}$ = $\sum_{i=1}^{n-1} \frac{(-1)^{r-1}r}{n \choose r}$, then S$_n$ = $\frac{n+1}{2n}$$S_{n-1}$ + $\frac{{(-1)}^n}{n}$, but could not go further.