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.

  • 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

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.


$$f(x)=\sum_{n=0}^{\infty}a_nx^n=a_0+a_1x+a_2x^2+...+a_nx^n+...$$ 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.


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