For a polynomial P(z) of degree $n\geq2$, show that there exists some $R_{1}>0$ such that for $R>R_{1}$ it holds that: $$\int_{C_{R}}\frac{1}{P(z)}dz=0$$ where $C_{R}$ is a circle of radius R with the centre $0$.
I am already aware of the fact that $$\lim_{R\to \infty}\int_{C_{R}}\frac{1}{P(z)}dz=0$$ and how to go about proving it but I'm not sure how to evaluate the first integral written above without using any limit. Would I maybe use a similar approach as was used to show the latter integral?
If $R$ and $S$ are greater than the absolute value of every root of $P$, then, by Cauchy's theorem$$\oint_{C_R}\frac1{P(z)}\,\mathrm dz=\oint_{C_S}\frac1{P(z)}\,\mathrm dz.$$This, together with the fact that that limit that you mentioned is $0$; is enogh to prove that $\oint_{C_R}\frac1{P(z)}\,\mathrm dz$ is $0$ if $R$ is large enough.