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

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 $$n=m+2$$.