1
$\begingroup$

I have the following statement:

If $\sum_{n=0}^{\infty}a_{n}x^n $ converges for $|x| < R$ , then $\sum_{n=0}^{\infty}na_{n}x^n $ converges for $|x| < R$ as well.

I couldn't find a counterexample, so I guess this is true , but I would like to get some hints for how to prove it.

$\endgroup$
  • 1
    $\begingroup$ Root test and $n^{1/n}→ 1$, if I recall correctly. $\endgroup$ – Calvin Khor May 28 '16 at 23:06
  • $\begingroup$ So u mean by doing the root test for $\sum_{n=0}^{\infty}na_{n}x^n $ ? $\endgroup$ – GeorgeB May 28 '16 at 23:11
  • $\begingroup$ yeah, and also for the original series (for which you already know converges on $|x|<R$). $\endgroup$ – Calvin Khor May 28 '16 at 23:12
2
$\begingroup$

Given an $x$ with $|x|<R$ choose $p$, $q$ such that $|x|<p<q<R$. Since $\sum_n a_nq^n$ converges there is an $M>0$ such that $|a_n|q^n\leq M$ for all $n\geq1$. Now we have $$|n\,a_n x^n|=\bigl(|x|/ p\bigr)^n\cdot n\> \bigl(p/q\bigr)^n\cdot |a_n|q^n\ .$$ Here $|x|/ p=:c<1$, and $\lim_{n\to\infty}n \bigl(p/q\bigr)^n=0$, hence $\>n \bigl(p/q\bigr)^n\leq M'$ for all $n\geq1$. This implies $$|n\,a_n x^n|\leq MM' c^n\qquad(n\geq1)\ ,$$ which is sufficient.

$\endgroup$
0
$\begingroup$

$$f(x)=\sum_{n=0}^{\infty}a_nx^n=a_0+a_1x+a_2x^2+...+a_nx^n+...$$ According to this $$f'(x)=a_1+2a_2x+...+na_nx^{n-1}+...$$ And according to the very same article:

Both of these series have the same radius of convergence as the original one.

And here is a rather technical paper.

$$xf'(x)=a_1x+2a_2x^2+...+na_nx^n+...$$ $$a_0+xf'(x)=a_0+a_1x+2a_2x^2+...+na_nx^n+...$$ $$a_0+xf'(x)=\sum_{n=0}^{\infty}na_nx^n$$

This shows the convergence, with the same radius as the original power series.

$\endgroup$

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.