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Let $$\sum_{n=0}^{\infty} a_nx^n$$ be a power series with convergence radius $R$ and $$\sum_{n=1}^{\infty} na_nx^{n-1}$$ the power series one gets by 'term-by-term differentiation' with the convergence radius $R'$.

I need to prove that $R = R'$.

My approach was to first prove $R \leq R'$ by assuming $R'< R$ and stating that there is an $r$ in between the two so that the power series is convergent and the derivative is not, which then means that $\lim_{n \to \infty} a_nr^n = 0$ while $\lim_{n \to \infty} na_nr^{n-1} \neq 0$ which is a contradiction(as I can split the limit up).

To prove that $R' \leq R$, I wanted to prove that $$|a_0|+ r\sum_{n=1}^{\infty} na_nx^{n-1}$$ is a majorant of $$\sum_{n=0}^{\infty} a_nx^n$$ and therefore the convergence radius is the same.

Is that prove valid or did I overlook something?

maths_xx
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    Note that the general term tending to $0$ is not equivalent to the convergence of the series. Specifically with power series it is true that if $|x_0|$ is strictly larger than the radius of convergence then the general term will not even tend to $0$ at the point $x_0$, but is it what you actually wanted to use here? In any case, there is a formula for the radius of convergence, and you can use it to prove that the radii are the same. – Mark Apr 04 '24 at 12:39
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    Recall that $R=\dfrac{1}{\limsup_{n\to\infty}|a_n|^{1/n}}$. – Alexander Burstein Apr 04 '24 at 12:41

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We know the radius of convergence of a series $\displaystyle\sum_{n\geq 0}a_nx^n$ is $$R=\dfrac{1}{\underset{n\to\infty}{\lim}\left|\dfrac{a_{n+1}}{a_n}\right|}$$ If we differentiate the series we get $$\frac{\mathrm d}{\mathrm dx}\displaystyle\sum_{n\geq 0}a_nx^n=\sum_{n\geq 0}na_nx^{n-1}=\sum_{n\geq 1}na_nx^{n-1}=\sum_{n\geq 0}(n+1)a_{n+1}x^{n}=\sum_{n\geq 0}b_nx^{n},$$ and applying the radius formula we have $$\begin{aligned} R'&=\dfrac{1}{\underset{n\to\infty}{\lim}\left|\dfrac{b_{n+1}}{b_n}\right|}\\ &=\dfrac{1}{\underset{n\to\infty}{\lim}\left|\dfrac{(n+2)a_{n+2}}{(n+1)a_{n+1}}\right|}\\ &=\dfrac{1}{\underbrace{\underset{n\to\infty}{\lim}\left|\dfrac{n+2}{n+1}\right|}_1\underset{n\to\infty}{\lim}\left|\dfrac{a_{n+2}}{a_{n+1}}\right|}\overset{n\to n-1}{=}\dfrac{1}{\underset{n\to\infty}{\lim}\left|\dfrac{a_{n+1}}{a_n}\right|}=R \end{aligned}$$

Hug de Roda
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  • I am unaware of the formula for the radius of convergence you use, nor am I sure such a limit always exists. – John Frank Sep 25 '24 at 06:59