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This is mechanized in Maple:
binomial(m+n, n)*(sum(binomial(m, k)n(-1)^k/(n+k), k = 0 .. m))
binomial(m+n, n)*(sum(binomial(m, k)*n*(-1)^k/(n+k), k = 0 .. m));
$${\frac {{m+n\choose n}}{{m+n\choose m}}}$$ PS. Please, don't change my answer.
$${\frac {{m+n\choose n}}{{m+n\choose m}}}$$