Review : Power Series — Question 3

Question 3 – Solution

Strategy. Differentiation and integration preserve the radius but can alter endpoints; continuity and the mean value theorem settle endpoint slopes.

Step 1: Check the original series. The ratio test gives $R=1$. At $x=1$ and $x=-1$, the absolute values sum to $\sum 1/n^2<\infty$. Thus $\boxed{I_F=[-1,1]}$, absolutely throughout. For $|x|>1$ the terms do not tend to zero.

Step 2: Differentiate and integrate inside the radius. For $|x|<1$, $$\boxed{F'(x)=\sum_{n=1}^{\infty}\frac{x^{n-1}}n,\qquad G(x)=\sum_{n=1}^{\infty}\frac{x^{n+1}}{n^2(n+1)}.}$$ The constant in $G$ is zero because $G(0)=0$. Both new radii are $1$. For the derivative series, $x=1$ gives harmonic divergence; $x=-1$ gives conditional alternating convergence. Hence $I_{F'}=[-1,1)$. The integrated series converges absolutely at both endpoints, so $I_G=[-1,1]$. Here these interval labels refer to the displayed series, not an assumed endpoint differentiation rule.

Step 3: Sum the derivative in the open interval. Using $\sum_{n\ge 1}x^n/n=-\log(1-x)$ gives $$F'(x)=\frac{-\log(1-x)}x\quad(0<|x|<1),\qquad F'(0)=1.$$ The value at zero follows from the constant term and also from the limit. Consequently $\lim_{x\to-1^+}F'(x)=\log 2$, whereas $\lim_{x\to 1^-}F'(x)=+\infty$.

Step 4: Justify what happens at the boundary. Since $|x^n/n^2|\le 1/n^2$ on $[-1,1]$, the uniform majorant test gives uniform convergence there and continuity of $F$. Apply the mean value theorem between an interior point and either endpoint. The intervening derivative then tends to the corresponding limit found above. Therefore $$\boxed{F'_+(-1)=\log 2,\qquad \lim_{x\to 1^-}\frac{F(1)-F(x)}{1-x}=+\infty.}$$ So the finite value $F(1)$ has no finite left derivative. Endpoint behavior must be proved separately even when the power-series radius is unchanged.