I've been doing some fiddling around with probabilities and came across an interesting equation that I'd like to prove:

$$ (N-1)\sum_{n=2}^{N} (-1)^{n} {N-2\choose n-2} \frac{1}{n} = \frac{1}{N} $$

I've tried a proof by induction, but I get to a point where I need to prove another equation of nearly the same form, and therefore run into the same problem (though I haven't tried proving that equation by induction). That equation is:

$$ N\sum_{n=2}^{N} (-1)^{n} {N-1\choose n-2} \frac{1}{n} = \frac{(-1)^{N}N+1}{N+1} $$

I could be approaching proof by induction the wrong way. Any help would be appreciated.


Hint: $\sum\limits_{m=0}^{N-2}(-1)^{N-2-m} \binom {N-2} {m} x^{m}=(1-x)^{N-2}$. Multiply by $x$ and integrate from $0$ to $1$. Then change the variable from $m$ to $n=m+2$.


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