For f(x) equal to the sum from n = 1 to infinity of xⁿ⁻¹/n, decide whether f is differentiable at x = 1 and at x = −1.

Step-by-step solution

1. 1

   Find the radius of convergence and a closed form. The coefficients are $1/n$, so the radius is $1$. For $0<|x|<1$, factor out $1/x$ and use the logarithm series:

   $f(x)=\frac1x\sum_{n=1}^{\infty}\frac{x^{n}}{n}=-\frac{\ln(1-x)}{x},$

   with the removable value $f(0)=1$. Having a closed form is what makes the endpoint questions answerable, since the differentiated series behaves badly there.

2. 2

   Settle $x=1$ by checking convergence first. At $x=1$ the series becomes $\sum_{n\ge1}\frac1n$, the harmonic series, which diverges. So $f(1)$ does not exist as a number, and the question of differentiability at $x=1$ does not even arise. The closed form agrees: $-\ln(1-x)/x\to+\infty$ as $x\to1^{-}$.

3. 3

   Settle the value at $x=-1$. Here the series is $\sum_{n\ge1}\frac{(-1)^{n-1}}{n}$, the alternating harmonic series, convergent by Leibniz's test with sum

   $f(-1)=\ln 2\approx 0.6931472.$

   Abel's theorem guarantees this equals $\lim_{x\to-1^{+}}\bigl(-\ln(1-x)/x\bigr)=-\ln 2/(-1)=\ln 2$, so the closed form extends continuously to $x=-1$.

4. 4

   Differentiate the closed form and evaluate at $-1$. From $f(x)=-\ln(1-x)\cdot x^{-1}$,

   $f'(x)=\frac{1}{x(1-x)}+\frac{\ln(1-x)}{x^{2}},$

   so

   $f'(-1)=\frac{1}{(-1)(2)}+\frac{\ln 2}{1}=\ln 2-\frac12.$

   Because $1-x=2>0$ and $x=-1\ne0$, the closed form is real-analytic on a whole neighbourhood of $-1$, which is why this one-sided derivative genuinely exists.

5. 5

   Note the trap and check numerically. Differentiating the series term by term gives $\sum_{n\ge1}\frac{(n-1)x^{n-2}}{n}$, whose terms at $x=-1$ have magnitude $(n-1)/n\to1$ and therefore do not tend to $0$ — the differentiated series diverges at $x=-1$. That does not contradict the result: it only shows term-by-term differentiation is invalid at the endpoint, while the function itself is still differentiable there. Numerically, $\bigl(f(-1+10^{-10})-f(-1)\bigr)/10^{-10}=0.1931472$ and $\ln 2-\tfrac12=0.1931472$.