I have been given a problem as part of a practice set for a problem solving class. Here, I am asked to prove the following equality (for positive integers $m$ and $n$):
$\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{k+m+1}=\sum_{k=0}^{m}(-1)^k\binom{m}{k}\frac{1}{k+n+1}$
I have tried to form a counting argument,though that doesn't seem to be working too well. The theme of these problems is that they can be re-written as simpler problems, although I am just not sure how to do that here.
Some other things I tried include rewriting each sum as the difference of two other sums (ie putting the positive elements into one sum and the negatives into another) but that doesn't seem to simplify it at all...
Any help is much appreciated
A nice technique for this kind of thing is to involve integrals: $$\eqalign{\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{1}{k+m+1} &=\sum_{k=0}^{n}(-1)^k\binom{n}{k}\int_0^1x^{k+m}\,dx\cr &=\int_0^1\sum_{k=0}^{n}(-1)^k\binom{n}{k}x^{k+m}\,dx\cr &=\int_0^1 x^m(1-x)^n\,dx\cr &=\int_0^1 y^n(1-y)^m\,dy\qquad\hbox{(substitute $y=1-x$)}\cr &=\cdots\qquad\hbox{(similar calculation, I'll leave it to you)}\cr &=\sum_{k=0}^{m}(-1)^k\binom{m}{k}\frac{1}{k+n+1}\ .\cr}$$