Let $a_n=\frac {1}{3^n}$ if $n$ is prime and $a_n=\frac {1}{4^n}$ if $n$ is not prime.

Question

I am stuck on the following problem that says:

Let $a_n=\frac {1}{3^n}$ if $n$ is prime and $a_n=\frac {1}{4^n}$ if $n$ is not prime. Then I have to find the radius of convergence of the power series $$\sum_{n=1}^{\infty} a_nx^n$$ ?

I know that for the power series, $$f(x) = \sum_{k=0}^{\infty} a_k x^k$$ the radius of convergence is given by $R$ such that $$\limsup_{n \to \infty} \left \vert a_n \right \vert^{1/n} R = 1$$ i.e. $$R = \dfrac1{\limsup_{n \to \infty} \left \vert a_n \right \vert^{1/n}}$$. Now I have two different values of $a_n$ depending on whether $n$ is prime or not prime. Now, if $n$ is prime then $R=3$ and otherwise $R=4$.

EDIT: Now as @Pambos has suggested that "Note that $\limsup x_k$ is the $\sup$ of the set of limit points of $(x_k)_{k\in\mathbb N}$. ".So I guess that the required radius of convergence of the given series is $4$.Am I going in the right direction? Thanks in advance for your time.

Answer 1

You want to calculate $\limsup x_k$ of a sequence $(x_k)_{k\in\mathbb N}$.

Note that $\limsup x_k$ is the $\sup$ of the set of limit points of $(x_k)_{k\in\mathbb N}$. Also if you partition $\mathbb N$ as $S_1\cup S_2$ (in your case primes $\cup$ nonprimes) the set of limit points of $(x_k)_{k\in\mathbb N}$ is limit points of $(x_k)_{k\in\mathbb s_1} \cup$ limit points of $(x_k)_{k\in\mathbb s_2}$.