Domain of convergence (when not in the correct form) Radius of convergence is for power series but how does one go about computing the radius of convergence of the infinite sum 
$$\sum_{k=1}^{\infty} \frac{k}{k+1}\left(\frac{2x+1}{x}\right)^k\ ?$$
Can you find the radius $R$ directly by the standard $1/\limsup$ formula (or the Ratio Test) or do you have to make some kind of substitution to get it into the right form? I got the domain of convergence $(-1, -1/3)$ when I used the Ratio Test.
If you do need a substitution, give me a hint of how to go about that. If I am okay computing $R$ correctly with Ratio or Root Test, give me a hint why that's acceptable (even though $(2x+1)/x$ isn't a term of a polynomial).
 A: You can use the standard formulas, since for fixed $x$, you get convergence or divergence of the series. Using the ratio test, one gets that 
$$
\left| \frac{a_{k+1}}{a_k} \right| = \frac{(k+1)^2}{k(k+2)} \left| \frac{2x+1}{x} \right| \to L < 1
$$
if and only if $|2x+1| < |x|$ with $x \neq 0$ (after computing). This gives you the interval $(-1,-1/3)$ indeed.
Hope that helps,
A: Yes, the Ratio Test will work. After forming and simplifying $\left|\frac{x_{n+1}}{x_n}\right|$, it will (typically) involve the variable $x$. Take the limit as $n\to\infty$ and then figure out what $x's$ will make the limit $L<1$. Convergence for the case when $L=1$ has to be decided some other way.
The idea of radius of convergence applies to any infinite series involving a variable, $x$ for example.
A: If you set $y=\frac{2x+1}{x}$, your series rewrites:
$$\tag{1} \sum_{k=1}^\infty \frac{k}{k+1}\ y^k$$
which is a power series.
The radius of convergence of (1) equals $1$, and (1) converges also for $y=-1$; therefore the convergence set of (1) is $[-1,1[$, i.e.:
$$\tag{2} -1\leq y<1$$
Finally, you can return to your original variable setting $y=\frac{2x+1}{x}$ in (2); solving (2) w.r.t. $x$ gives the convergence set of your original series.
In fact your original series converges if and only if $x$ solves:
$$\begin{cases} \frac{2x+1}{x}\geq -1 \\ \frac{2x+1}{x}<1 \end{cases} \quad \Leftrightarrow \quad \begin{cases} \frac{3x+1}{x}\geq 0 \\ \frac{x+1}{x}<0 \end{cases} $$
hence iff $x\in ]-1 ,-1/3]$.
