I have a series $\sum_{n=0}^{\infty}a_nx^n$ with $1/n^2\leq a_n\leq 1/n$ and I need to find its radius of convergence, R.
Here's what I've done:
Since $1/n^2\leq a_n\leq 1/n$, we have $\sum_{n=1}^{\infty}x^n/n^2\leq \sum_{n=0}^{\infty}a_nx^n \leq \sum_{n=1}^{\infty}x^n/n$. Additionally, since $\lim_{n\to \infty}|(n+1)^2/n^2|=1$, the series $\sum_{n=1}^{\infty}x^n/n^2$ has a radius of convergence of 1 and thus R is at least 1. Similarly since $\lim_{n\to \infty}|(n+1/n|=1$, the radius of convergence of the series $\sum_{n=1}^{\infty}x^n/n$ is also 1, and thus R is 1 at most. This implies that R is exactly 1.
Does this seem fine? Do I need to specify why R is between the radii of convergence of the two other series, or is this clear enough? Thanks in advance!
As noted by @GReyes , your comparisons of sums are valid only for $x\ge0.$ For your proof to be complete, just argue that this is sufficient ($R$ is determined by the fact that the series is convergent for every $x\in[0,R)$ and divergent for every $x>R$).
But you could as well derive directly the result in the following way: $\sqrt[n]{\frac1n}=e^{-\frac{\ln n}n}\to e^0=1$ and $\sqrt[n]{\frac1{n^2}}=\left(\sqrt[n]{\frac1n}\right)^2\to1^2,$ hence (squeezing+Cauchy-Hadamard) $\frac1R=\lim\sqrt[n]{a_n}=1.$