I would like to prove that if $\sum_{n=0}^{\infty}a_{n}x^{n}$ converges on $(-R,R)$, then $\sum_{n=0}^{\infty}\frac{1}{n+1}a_{n}x^{n+1}$, $x\in(-R,R)$ converges on $(-R,R)$. I think it is often done by showing that the radius of convergence is the same. I was wondering if the fowlling is a valid argument or if not where I went wrong.

Let $x\in(-R,R)$. Because $\sum_{n=0}^{\infty}a_{n}x^{n}$ converges we know that $lim_{n\rightarrow\infty}a_n x^n=0$. Since every convergent sequence is bounded we have $|a_nx^n|\leq M$.

Therefore $|\frac{1}{n+1}a_{n}x^{n+1}|=|a_{n}x^n| |\frac{x}{n+1}|\leq M\frac{|x|}{n+1}\leq M|R|$. Now I would like to apply the Weierstrass–M–Test to conclude that the series converges uniformly on $(-R,R)$ by the Weierstrass–M–Test.

Thanks a lot in advance!

  • 1
    $\begingroup$ Hint: calculate $\limsup \sqrt[n]{|a_n|/(n+1)}$ and use Cauchy theorem $\endgroup$ – Jakobian Jul 20 at 23:06

That is not a valid argument because the series $\sum_{n=0}^\infty M\lvert R\rvert$ actually diverges (unless $M=0$), and therefore you did not prove that it follows from the Weierstrass $M$-test that your series converges.


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