Find the sum of x^(n-1)/n for n from 1 to infinity.

Step-by-step solution

1. 1

   Shift the exponent to reach a standard series. The familiar identity has $x^{n}$ on top, not $x^{n-1}$. For $x\neq 0$ pull one factor of $x$ out of the whole sum:

   $\sum_{n=1}^{\infty}\frac{x^{\,n-1}}{n}=\frac{1}{x}\sum_{n=1}^{\infty}\frac{x^{n}}{n}$

   The restriction $x\neq 0$ is introduced here and must be revisited later.

2. 2

   Use the logarithm series. Integrating the geometric series $\sum_{n\ge 0}t^{n}=\dfrac{1}{1-t}$ term by term from $0$ to $x$ gives

   $\sum_{n=1}^{\infty}\frac{x^{n}}{n}=-\ln(1-x),\qquad |x|<1$

3. 3

   Divide by $x$ to get the closed form.

   $\sum_{n=1}^{\infty}\frac{x^{\,n-1}}{n}=-\frac{\ln(1-x)}{x},\qquad 0<|x|<1$

4. 4

   Handle $x=0$ separately. The original series at $x=0$ is $\dfrac{0^{0}}{1}+0+0+\cdots$, whose first term is the ambiguous $0^{0}$; the closed form is likewise $\tfrac00$. Both are resolved by a limit:

   $\lim_{x\to 0}\frac{-\ln(1-x)}{x}=\lim_{x\to 0}\frac{1/(1-x)}{1}=1$

   by L Hopital. So $x=0$ is a removable singularity, and the natural value there is $1$.

5. 5

   State the complete answer.

   $\sum_{n=1}^{\infty}\frac{x^{\,n-1}}{n}=\begin{cases}-\dfrac{\ln(1-x)}{x},&0<|x|<1\\[4pt]1,&x=0\end{cases}$

   Numerically at $x=0.4$ a 400-term partial sum gives $1.27706405941$ and $-\ln(0.6)/0.4=1.27706405941$ ✓.