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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 :

$$g(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$$

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.

Proof.

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 $\displaystyle g'( x)$.

Proof.

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*}

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    $\begingroup$ 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

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Hint

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.)

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