# Show that this power series converges for |x|<2

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 $$\limsup \sqrt[n]{|b_n|}?$$
• What you need is that $n^{1/n} \to 1$. This shows that $n^ka_n$ has the same radius of convergence for any fixed $k > 0$. Sep 22, 2015 at 4:38
• Hi @martycohen, silly question but does this mean that $n^{k/n}$ -> 1? Thanks... Sep 22, 2015 at 4:54
• Probably does, since it's $(n^{1/n})^k$ -> $1^k$ = 1, right? Thanks, @martycohen ... Sep 22, 2015 at 4:56