Interval of convergence of $\sum^\infty_{n=1} \frac 1n (\frac{x-1}x)^n$ $$\sum^\infty_{n=1} \frac 1n \left(\frac{x-1}x\right)^n$$
Questions: How to find interval of convergence? And is sum of the series continuous on the interval of convergence?
I know how to use ratio test for power series, but it's kind of strange power series. Give me advice, please.
addition: I notice that $x \neq 0 \notin $ interval of convergence.
 A: Hint. Let $z=\frac{x-1}{x}$ and view it as a power series in $z$. Use what you know to find the interval convergence as a power series in $z$. Then solve for the interval of convergence in terms of $x$. 
A: For $y=(x-1)/x$ you get the power series $\sum \frac{1}{n}y^n$ it will converge in a set of the form $|y|<R$. Then you can solve the inequality $|(x-1)/x|<R$ to get your answer.
A: Substitute some variable in for $\frac{x-1}{x}$, for example, $y=\frac{x-1}{x}$ Then use the ratio test, as you know how to do, to determine the interval of convergence for this: 
$\sum_{n=1}^{\infty}\frac{y^n}{n}$
You will need to substitute $\frac{x-1}{x}$ back in for y when you have completed applying the test. Embedded expressions like that usually call for some kind of substitution.
It can be done like this: 
$\lim_{n\rightarrow \infty}\left | \frac{\frac{y^{n+1}}{n+1}}{\frac{y^n}{n}} \right |= \lim_{n\rightarrow \infty}\left | \left ( \frac{n}{y^n}\left ( \frac{y^{n+1}}{n+1} \right ) \right ) \right |=\lim_{n\rightarrow \infty}\left | \frac{ny}{n+1} \right |=\left | y \right |\lim_{n\rightarrow \infty}\frac{n}{n+1}$.
Since the last limit above evaluates to 1, you are simply looking at $\left |y \right |$ < 1. (I set it less than one, because the ratio test definitely converges when the limit is less than one.) We can solve this by substituting our original value of $y$, which was $(x+1)/x$, back into the absolute value inequality and find the values of x that make the inequality true. 
