# Proving the claim that if a power series converges then the differentiated series also converges

I am self-learning Real Analysis from the text Understanding Analysis by Stephen Abbott. I am interested to prove the below theorem. Do you have a hint/clue for part (b) of this exercise problem, without revealing the entire solution?

Theorem. If $$\sum_{n=0}^{\infty}a_n x^n$$ converges for all $$x\in(-R,R)$$, then the differentiated series $$\sum_{n=1}^{\infty}na_n x^{n-1}$$ also converges at each $$x \in (-R,R)$$ as well. Consequently, the convergence is uniform on compact sets contained in $$(-R,R)$$.

[Abbott 6.5.5] (a) If $$s$$ satisfies $$0 < s < 1$$, show that $$ns^{n-1}$$ is bounded for all $$n \geq 1$$.

Proof.

Let $$\displaystyle C=\frac{1}{s}$$. Then, $$\displaystyle C >1$$. Consider:

$$\begin{equation*} \lim _{n\rightarrow \infty } ns^{n-1} =\lim _{n\rightarrow \infty }\frac{n}{C^{n-1}} \end{equation*}$$ This is of the form $$\displaystyle \frac{\infty }{\infty }$$. Applying the L'hopital's rule:

$$\begin{equation*} \lim _{n\rightarrow \infty } ns^{n-1} =\lim _{n\rightarrow \infty }\frac{n}{C^{n-1}} =\lim _{n\rightarrow \infty }\frac{1}{C^{n-1}\log C} =0 \end{equation*}$$

This is also apparent by the fact, that an exponential term grows much faster than a polynomial term.

Since convergent sequences are bounded, it implies that $$\displaystyle ns^{n-1}$$ is a bounded sequence.

(b) Given an arbitrary $$x \in (-R,R)$$, pick $$t$$ to satisfy $$|x|. Use this start to construct a proof for the theorem 6.5.6.

Proof.

Fix $$\displaystyle x_{0} \in ( -R,R)$$ and pick $$\displaystyle t$$ such that $$\displaystyle 0\leq |x_{0} |< t< R$$.

Define

$$\begin{equation*} f_{n}'( x) \ =\ na_{n} x^{n-1} \end{equation*}$$

We have:

$$\begin{equation*} 0\leq |f_{n}'(x_{0}) |=|na_{n} x_{0}^{n-1} |\leq n\left(\frac{t}{R}\right)^{n-1} \cdotp |a_{n} |\ R^{n-1} \end{equation*}$$

Now, $$n(t/R)^{n-1}$$ is bounded, so there exists $$C$$, for all $$n\in \mathbf{N}$$ such that $$|n(t/R)^{n-1}|\leq C$$. But, how do I handle $$|a_n|R^{n-1}$$. We only have convergence on $$(-R,R)$$ and not at the endpoint $$x = R$$.

• “We only have convergence on $(-R,R)$ ... ” – yes that is that the theorem states – “... and not at the endpoint $x = R$.” – the theorem does not make any claim about convergence at the endpoints. Dec 24, 2022 at 11:10

My hint is that nobody is making you write $$R$$ in two places in your last math line $$|na_{n} x_{0}^{n-1} |\leq n\left(\frac{t}{R}\right)^{n-1} \cdotp |a_{n} |\ R^{n-1}.$$ The same inequality would hold if you instead wrote $$s$$ in those places, where you choose $$s$$ satisfying $$t.
You don't actually need a separate variable $$s$$. Instead you could just write $$|na_{n} x_{0}^{n-1} |\leq n\left(\frac{|x_0|}{t}\right)^{n-1} \cdotp |a_{n} |\ t^{n-1}.$$ Either way, once you've set that up, the term $$|a_n| R^{n-1}$$ that you were worried about are now replaced by something like $$|a_n| t^{n-1}$$ which converges just fine since $$t$$ is inside the radius of convergence.
• Okay. Basically, for an arbitrary $x \in (-R,R)$, you can always pick $t$ satisfying $0 \leq |x| < t< R$, so $|x|/t < 1$. Then, you have $|na_n x^{n-1}| \leq n(|x|/t)^{n-1} |a_n| t^{n-1}$, and we know that $\sum a_n x^{n}$ is absolutely convergent at $t < R$. I need to complete the argument. Dec 24, 2022 at 11:44
• how do I take care of the fact that we have the term $|a_n|t^{n-1}$ instead of $|a_n|t^n$ on the right hand side of the above inequality? I know that, $\sum a_n x^n$ converges. But, we have no knowledge of the series $\sum a_n x^{n-1}$. Dec 25, 2022 at 8:09
• @Quasar As you said, we know $\sum_{n=1}^\infty |a_n| t^n$ converges. If you divide that series by $t$ you get $\sum_{n=1}^\infty |a_n| t^{n-1}$, so the latter series also converges. Dec 25, 2022 at 11:17
• It makes sense now. If $\sum_{k=1}^{\infty} a_k = A$, then $\sum_{k=1}^{\infty} c a_k = cA$ from the properties of infinite series of reals. Dec 27, 2022 at 5:54