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

Let $$(a_n),(b_n)$$ be sequences of positive real numbers such that $$na_n for all $$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

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?

• Yes fixed it Feb 20, 2019 at 8:07
• What do you think is the correct answer ? Feb 20, 2019 at 8:07

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.
• you mean $\sum a_n x^n$ and $\sum b_n x^n$ right? so the correct option is A, is that right? Feb 20, 2019 at 8:09