Suppose that $$A:=\sum a_nx^n< \infty $$
for |x|<2. Let $$|b_n|<n^2|a_n|$$
Show that $$B:=\sum b_n x^n < \infty$$
for |x|<2.
My work: we know that A converges uniformly inside the interval (-2,2). Differentiate term-by-term to get
$$A':=\sum na_nx^{n-1}< \infty $$
and we know that A' has the same radius of convergence as A.
Now, can I multiply A' by powers of x to get other convergent series? I think I can but not 100% sure, since the summands will be bigger.
Can I say that, based on the convergence of A', then
$$\sum na_nx^n< \infty $$
for |x|<2? I just simply multiplied A' by one power of x.
If that is valid and does not change the radius of convergence, then I do it once more to get
$$\sum n^2a_nx^{n-1}< \infty $$
implies
$$ \sum n^2a_nx^n< \infty $$
for |x|<2. (First differentiate term-by-term, then multiply by one power of x.)
Finally, using the bound $|b_n|<n^2|a_n|$, I want to use the comparison test and say that since
$$\sum|b_n||x|^n<\sum n^2|a_n||x|^n$$
then somehow $\sum|b_n||x|^n$ converges for |x|<2, which implies that
$\sum b_nx^n$ converges for |x|<2 as was to be shown.
The problem is that I don't know whether the absolute series $\sum n^2|a_n||x|^n$ converges. We know that absolute convergence implies convergence, but I don't think the converse is true.
How can I tweak my solution to get to the right answer?
Thanks,
Hint: Use the Cauchy-Hadamard Theorem. From the hypotheses, you may conclude that $$\limsup \sqrt[n]{|a_n|}\le 0.5$$
Now, what can you say about $$\limsup \sqrt[n]{|b_n|}?$$