1) Il suffit de remarquer que f(n+1)=(n^2+1)/(n+1)^2 *f(n).
2) Congruence 2.
3) f(x)=sqrt(x(1-x)) est connexe sur (0,1), sum_1^n f(x_i) =<n f((sum_1^n x_i)/n)=sqrt(n-1).
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Two things are infinite: the universe and human stupidity; and I'm not sure about the the universe