I need to show that f(z)=0 for all z \in D(0,2).
From the analyticity of f in D(o,2), I know by Cauchy's theorem it's integral in |z|<2 is zeros. And clearly the integrand has a pole at 1/(n+1) for each n. Do I have to use Identity theorem to prove the desired result. Any help is much appreciated.
Multiply both sides by $(n+1)/(2\pi i)$ to obtain $$ \frac{1}{2\pi i}\int_{|z|=1} \frac{f(z)}{z-\frac{1}{(n+1)}}\,dz = 0. $$ Then notice that by the Cauchy integral formula, the integral on the left is precisely $f(1/(n+1))$. So the zero set of $f(z)$ has an accumulation point inside $D(0,2)$, so it must be identically zero there.