Prove that $\sum_{r=1}^n (-1)^{r-1}(1+\frac{1}{2}+\frac{1}{3}+...\frac{1}{r}) \binom{n}{r} =\frac{1}{n}$. Where $\binom{n}{r}$ represents 'n choose r'.
I tried to simplify this expression by first trying to manipulate around the relation $\frac{\binom{n}{r}}{\binom{n}{r+1}}=\frac{n-r+1}{r} $.
However the above relation doesn't seem to help as $\binom{n}{r}$ is being divided by r and not multiplied wherever such terms appear. I also tried to expand the summation into multiple summations like:
$\sum_{r=1}^n (-1)^{r-1}\binom{n}{r}+\sum_{r=2}^n (-1)^{r-1}\frac{\binom{n}{r}}{2}+...\sum_{r=n}^n (-1)^{r-1}\frac{\binom{n}{r}}{n}$
and then wrote it in reverse as I saw the $(-1)^{r-1}$ and felt terms might cancel. In doing so, I am faced with another problem. Say normally I had two such binomial summations A,B where $A+B=0$, in writing the reverse, I do not get $A+B$ and rather terms like $\frac{A_2}{2}+\frac{B_2}{n-1}$, $\frac{A_3}{3}+\frac{B_3}{n-2}$ and such. I am unsure of how to proceed so please help me.