I am self-learning Real Analysis from the text Understanding Analysis by Stephen Abbott. Exercise problem 6.5.1 asks to find the domain of convergence of the power series :


Do you have clue/hint (without revealing the entire solution), on whether $g$ is defined for $|x|>1$? Also, hope my proof attempt checks out.

[Abbott 6.5.1] (a) Is $\displaystyle g$ defined on $\displaystyle ( -1,1) ?$ Is it continuous on this set? Is $\displaystyle g$ defined on $\displaystyle ( -1,1]$? Is it continuous on this set? What happens on $\displaystyle [ -1,1]$? Can the power series for $\displaystyle g( x)$ possibly converge for any other points $\displaystyle |x| >1$? Explain.


Fix $\displaystyle x_{0} \in ( -1,1)$. Define:

\begin{equation*} a_{n} =\frac{x_{0}^{n}}{n} \end{equation*} Then, $\displaystyle \sum _{n=1}^{\infty }( -1)^{n+1} a_{n}$ is an alternating series. We have, $\displaystyle a_{1} \geq a_{2} \geq \dotsc \geq 0$ and $\displaystyle ( a_{n})\rightarrow 0$. Hence, by the alternating series test, $\displaystyle \sum ( -1)^{n+1} a_{n}$ converges on $\displaystyle ( -1,1)$. Hence, $\displaystyle g$ is defined on $\displaystyle ( -1,1)$. Since a power series is continuous on it's domain of convergence, it is continuous on $\displaystyle ( -1,1)$.

Since $\displaystyle \sum _{n=1}^{\infty }\frac{( -1)^{n+1}}{n}$ is convergent, $\displaystyle g$ is defined at $\displaystyle x=1$. Hence, $\displaystyle g$ is continuous on $\displaystyle ( -1,1]$.

Since $\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}$ is divergent, $\displaystyle g$ is not defined at $\displaystyle x=-1$.

(b) For what values of $\displaystyle x$ is $\displaystyle g'( x)$ defined? Find a formula for $\displaystyle g'( x)$.


By the theorem on the convergence of a power-series, if a power series converges on $\displaystyle A$, it is continuous on $\displaystyle A$ and differentiable on all $\displaystyle ( -R,R) \subseteq A$. Thus, $\displaystyle g'$ is defined on $\displaystyle ( -1,1)$. Also, $\displaystyle g'$ is given by the term-by-term differentiation of $\displaystyle g$:

\begin{equation*} g'( x) =\sum _{n=1}^{\infty }( -1)^{n-1} x^{n-1} =1-x+x^{2} -x^{3} +\dotsc \end{equation*}

  • 1
    $\begingroup$ To converge, necessary condition for a series is that its term tend to $0$. This is not sufficient condition (as $\sum 1/n$ shows) but it is necessary. $\endgroup$
    – Salcio
    Dec 23, 2022 at 12:33

1 Answer 1



By Cauchy-Hadamard the radius of convergence is $r=\limsup_{n\to\infty}\dfrac 1{\lvert \dfrac {(-1)^{n+1}}n\rvert^{\frac1n}}=\limsup_{n\to\infty}n^{\frac 1n}$.

But as for $\lvert x\rvert \gt1,$ the general term $\dfrac {(-1)^{n+1}x^n}n\not\to0,$ so the above is overkill and the "divergence test" suffices. (It doesn't go to zero because exponential growth is faster than polynomial. You can always use L'hopital.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .