# Convergence of $\sum_{n=0}^{\infty}a_{n}x^{n}$ implies convergence of $\sum_{n=0}^{\infty}\frac{1}{n+1}a_{n}x^{n+1}$, $x\in(-R,R)$

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!

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

## 1 Answer

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.