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?



1 Answer 1


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, 2015 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, 2015 at 4:38
  • $\begingroup$ Hi @martycohen, silly question but does this mean that $n^{k/n}$ -> 1? Thanks... $\endgroup$
    – User001
    Sep 22, 2015 at 4:54
  • $\begingroup$ Probably does, since it's $(n^{1/n})^k$ -> $1^k$ = 1, right? Thanks, @martycohen ... $\endgroup$
    – User001
    Sep 22, 2015 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, 2015 at 5:20

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