# Show the power series $\sum_{n=1}^\infty na_nx^{n-1}$ diverges when $|x|>R$.

Suppose that the power series $$\sum_{n=1}^\infty a_nx^n$$ has radius of convergence $R$. Then the power series $$\sum_{n=1}^\infty na_nx^{n-1}$$ diverges if $|x|>R$.

Attempt: I can show that $\sum_{n=1}^\infty |na_nx^{n-1}|$ diverges by comparison test, but how do I show that $\sum_{n=1}^\infty na_nx^{n-1}$ diverges also?

• The comment of the other user about the limit comparison test that was deleted. – Michael L. Oct 10 '17 at 21:25
• @rtybase nope. This requires to prove the other direction also (direction of divergence, which is THE question for this post.) – 2ndaccount Oct 10 '17 at 23:02
• From $\sum a_n x^n < \infty, |x|<R \Leftrightarrow \sum na_n x^{n-1} < \infty, |x|<R$, then the problem of divergence of $\sum na_n x^{n-1}, |x|>R$ is equivalent with divergence of $\sum a_n x^n, |x|>R$. $$\Rightarrow$$ If $\sum a_n x^n < \infty, |x|<R$ then (en.wikipedia.org/wiki/…) $\sum na_n x^{n-1} < \infty, |x|<R$ - derivative is applied. $$\Leftarrow$$ $\sum na_n x^{n-1} < \infty, |x|<R$ then (same article) $\sum a_n x^n < \infty, |x|<R$ - integration is applied. – rtybase Oct 10 '17 at 23:18
• Yes, how do you prove that? Since I can prove the direction of convergence, here I'm asking for divergence. – 2ndaccount Oct 10 '17 at 23:20
• I updated my comment ... plus the logical trick of $p\iff q\equiv \neg p\iff \neg q$ – rtybase Oct 10 '17 at 23:23

Showing that it diverges for $|x|>R$ is equivalent to showing that the radius of convergence $R'$ of the second power series is $\leq R$.

But $\sum_{n=1}^{+\infty} |na_{n}x^{n-1}| = \frac{1}{x}\sum_{n=1}^{+\infty} |na_{n}x^{n}| \geq \frac{1}{x}\sum_{n=1}^{+\infty} |a_{n}x^{n}|$ shows just that.

P.S.: in fact the other direction ($R' \geq R$) is more difficult to prove

• Is the inequality on the right true? – 2ndaccount Oct 10 '17 at 21:54
• It is certainly true if $a_n$ is nonnegative. – 2ndaccount Oct 10 '17 at 21:55
• Corrected, i meant to consider the absolute values... which is what you are looking at for power series. – Julien Oct 10 '17 at 21:57
• I've already showed that but I'm having trouble showing true without the absolutely value. – 2ndaccount Oct 10 '17 at 22:00
• If this is absolutely diverging then $R' \leq R$ which means that for $x > R$ the sum will diverge because its main term $(a_{n}x_{0}^{n})$ is not bounded (when $x_{0}>R$, as per Abel's lemma). – Julien Oct 10 '17 at 22:07

Given that $\sum_{n=1}^\infty a_nx^n$ has a radius of convergence $R$, then we have from the Root Test that

$$R=\frac{1}{\limsup_{n\to \infty}\sqrt[n]{|a_n|}}$$

Inasmuch as $\lim_{n\to \infty}\sqrt[n]{n}=1$, then the radius of convergence of the series $\sum_{n=1}^\infty na_nx^{n-1}$ is

$$\frac{1}{\limsup_{n\to \infty}\sqrt[n]{|na_n|}}=\frac{1}{\limsup_{n\to \infty}\sqrt[n]{|a_n|}}=R$$

as was to be shown!

• so you are getting the radius of convergence for both of them and show that those are the same. Would you say that the approach of using comparison test on the two series is just not going to work? – 2ndaccount Oct 10 '17 at 21:42