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 $n \geq 1$.


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 $|x|<t<R$. Use this start to construct a proof for the theorem 6.5.6.


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


\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 $x = R$.

  • 3
    $\begingroup$ “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. $\endgroup$
    – Martin R
    Dec 24, 2022 at 11:10

1 Answer 1


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 $t<s<R$.

More explicit hinting/solution follows:

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.

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    $\begingroup$ 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. $\endgroup$
    – Quasar
    Dec 24, 2022 at 11:44
  • $\begingroup$ 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 $\sum a_n x^{n-1}$. $\endgroup$
    – Quasar
    Dec 25, 2022 at 8:09
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    $\begingroup$ @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. $\endgroup$ Dec 25, 2022 at 11:17
  • $\begingroup$ 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. $\endgroup$
    – Quasar
    Dec 27, 2022 at 5:54

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