# Radius of convergence for series of $\frac{1}{x^2+\cos^2x}$

Given $$f(x) = \frac{1}{x^2+\cos^2(x)}$$ We can use the binomial expansion to rewrite $$f(x)$$ as $$f(x) = \sum_{k=0}^\infty {-1\choose{k}}\frac{\cos^{2k}x}{x^{2k+2}}$$ Or, in other terms $$f(x) = \sum_{k=0}^\infty \frac{(-1)^k \cos^{2k}x}{x^{2k+2}}$$ If I remember correctly the formula for R.O.C is: $$(x+y)^n \Longrightarrow \, -1<\frac{x}{y} < 1$$ How would I go about solving this for: $$-x^2 < \cos^2x < x^2$$ Thanks.

• ROC is for power series and your series is not a power series. Commented Feb 15, 2021 at 8:43

The term radius of convergence is reserved for power series, such as $$\sum_n a_nx^n$$. Your series does not fit this format. However, it is a geometric series in $$\cos^2 x/x^2$$. It will converge if and only if $$\left\lvert \frac {\cos^2 x}{x^2} \right \rvert < 1.$$ At a minimum this will be the case whenever $$|x| > 1$$ since (for real values of $$x$$) $$|\cos x| < 1$$. But in fact, still concentrating on real $$x$$, for $$0 < x < 1$$, the ratio $$\cos^2x/x^2$$ is decreasing and greater than $$1$$ for small $$x$$. Therefore the inequality is true for all larger $$x$$ once $$x > \theta$$ where $$\theta$$ is the solution to $$\cos \theta = \theta$$, about $$0.7$$. Because each term is an even function, the sign of $$x$$ is unimportant and convergence of your sum arises for all $$|x|> \theta$$, which is the solution to the inequality you have written.