The power series $\sum \frac{(x-b)^n}{na^n}$ with a,b>0?

Question

The power series $\sum \frac{(x-b)^n}{na^n}$ with a,b>0 ?
How do i show for which x the series is conditionally convergent? Do i have to express in terms of a and b. or something like that?

Answer 1

By the root test this converges only when $$\lim_{n\to\infty} \left|\frac{x-b}{\sqrt[n]{n} a}\right|\lt 1$$ $$b-a\lt x \lt b+a$$

Answer 2

$$A_n=\frac{(x-b)^n}{na^n}\implies \frac{A_{n+1}}{A_n}=\frac n{n+1} \frac{x-b}{a}\sim \frac{x-b}{a}$$

If you let $z=\frac{x-b}{a}$, $$\sum_{n=1}^\infty \frac{(x-b)^n}{na^n}=\sum_{n=1}^\infty \frac{z^n} n $$ is a well know series.

Answer 3

The series converges if $|x-b| <a$ and diverges if $|x-b| >a$. (use ratio test). If $|x-b|=a$ the either $x-b =a$ or $x-b=-a$. In the first case we get the divergent series $\sum \frac 1 n$ and in the second case we get that conditionally convergent series $\sum (-1)^{n} \frac 1 n$.