Let $(a_n),(b_n) $ be sequences of positive real numbers such that $na_n<b_n<n^2a_n$

Question

Let $(a_n),(b_n) $ be sequences of positive real numbers such that $na_n<b_n<n^2a_n$ for all $n\geq2$. If the radius of convergence of $\sum a_n x^n$ is $4$, then $\sum b_n x^n$

A)converges for all $|x|<2$

B)converges for all $|x|>2$

C)does not converges for any $x$ with $|x|<2$

D)does not converges for any $x$ with $|x|>2$

I tried to solve this by taking $a_n = \frac{1}{(2n)!}$ so that radius of convergence of $\sum a_n x^n$ is $4$

Then I took $b_n$ = $n^{1.5}a_n$ so it satisfies $na_n<b_n<n^2a_n$

Now radius of convergence of $\sum b_n x^n = \sum n^{1.5}a_n$ is still $4$, so I think option A must be right ?

is this correct?

Answer 1

Use squeeze theorm and the fact that $\lim n^{1/n}=\lim (n^{2})^{1/n}=1$ to see that $\sum a_n x^{n}$ and $\sum b_n x^{n}$ have the same radius of convergence. A) is true. B),C) and D)are all false.