Radius of convergence for series of $\sum_{n=0}^{\infty}\big(2+(-1)^n\big)x^n.$

Question

I am asked to find the radius of convergence for that series $$\sum_{n=0}^{\infty}\big(2+(-1)^n\big)x^n.$$ It is obvious that i cant just find the limit of the fraction of two consecutive terms. Can i split the series into two different ones and work with those two radiuses ? Here it happens that both of them are 1 but does that imply that the radius of the overall series is also 1 ?

Answer 1

Yes, it does (in this case, at least). Let $a_n=2+(-1)^n$, then, for each $n\in\mathbb N$,$$1\leqslant\sqrt[n]{a_n}\leqslant\sqrt[n]3$$and therefore $\lim_{n\to\infty}\sqrt[n]{a_n}$. So, yes, the radius of convergence is $1$.

Answer 2

Note the sequence $\{a_n\}$ generated by coefficients of the series is of the form $$a_n=\begin{cases} 3\text{ if } n\text{ is even,}\\ 1\text{ if } n\text{ is odd.}\end{cases}$$ So $$\limsup_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{m\to\infty}\sqrt[2m]{|a_{2m}|}=\lim_{m\to\infty}\sqrt[2m]{3}=1.$$ Hence radius of convergence is $$\frac{1}{\limsup_{n\to\infty}\sqrt[n]{|a_n|}}=1.$$