# Domain of convergence of the power series $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$

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 $$\displaystyle g$$:

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

• 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. Dec 23, 2022 at 12:33

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.)