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

$$\sum n^2a_nx^{n-1}< \infty $$


$$ \sum n^2a_nx^n< \infty $$

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?



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 $$\limsup \sqrt[n]{|b_n|}?$$

  • $\begingroup$ Hi @vadim123, shouldn't it be an equality instead of the weak-inequality? $\endgroup$
    – User001
    Sep 22 '15 at 4:24
  • 1
    $\begingroup$ 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 $k > 0$. $\endgroup$ Sep 22 '15 at 4:38
  • $\begingroup$ Hi @martycohen, silly question but does this mean that $n^{k/n}$ -> 1? Thanks... $\endgroup$
    – User001
    Sep 22 '15 at 4:54
  • $\begingroup$ Probably does, since it's $(n^{1/n})^k$ -> $1^k$ = 1, right? Thanks, @martycohen ... $\endgroup$
    – User001
    Sep 22 '15 at 4:56
  • 1
    $\begingroup$ Yep. I intended for that to be implied for the reason stated. It's a slam dunk. $\endgroup$ Sep 22 '15 at 5:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.