What is the radius of convergence of the power series $\sum_{n=1}^{\infty}\frac{n}{n+1}\left(\frac{2x+1}{x}\right)^n$ I want to finf the radius of convergence of this power series, so what I did is that I calculated $\displaystyle\lim_{n \to \infty}\left|\dfrac{a_{n+1}}{a_n}\right|$ and shown it to be less than $1$ for the power series to be convergent. That is
$$\displaystyle\lim_{n\to \infty}\left|\dfrac{n+1}{n+2}\cdot\dfrac{n+1}{n}\cdot\left(2+\dfrac{1}{x}\right) \right| < 1$$
Solving for $x$ I got $-3 < \dfrac{1}{x} < -1 \implies -1 < x < -\dfrac{1}{3}$. From here I calculated the length of half of the interval and that was it, the radius of convergence came out to be $\dfrac{1}{3}$. But the answer for this question was $1$ and I don't trust the source of this answer. Is my solution correct?
Edit: The question was in my test as it is in the title word to word. If it meant for what values of $x$ does the series converge or what is the radius of convergence w.r.t $x$ (if it even exists) then would I be correct? If it was w.r.t to $\left( \dfrac{2x+1}{x}\right)$ then I know the radius of convergence would be $1$.
 A: The difficulty, as indicated in the comments, is you were given a "power series" but the quantity raised to a power is an algebraic expression, not a single variable. So do we seek the radius of convergence with respect to $x$ or with respect to $(2x+1)/x$? You do indeed get $1/3$ with the former interpretation versus $1$ with the latter.
As you found, you get different answers based on these two plausible but different interpretations. Therefore unless the proper variable (just $x$ or the full expression $(2x+1)/x$) is specified for measuring the radius, the problem is really not well-posed.
Lemons to lemonade (and logarithms)
We could also exploit this ambiguity to our advantage. Consider the familiar Maclaurin series
$\ln (1+y) =y-\dfrac{y^2}2+\dfrac{y^3}3-\dfrac{y^4}4+\dfrac{y^5}5-...$
We can of course reverse the sign of $y$ to get:
$\ln (1-y) =-y-\dfrac{y^2}2-\dfrac{y^3}3-\dfrac{y^4}4-\dfrac{y^5}5-...$
And subtract to get the logarithm of the quotient:
$\ln\left(\dfrac{1+y}{1-y}\right)=2\left(y+\dfrac{y^3}3+\dfrac{y^5}5+...\right)$
Now suppose we define $x=(1+y)/(1-y)$, from which we may obtain $y=(x-1)/(x+1)$. Thus
$\ln x=2\left(y+\dfrac{y^3}3 +\dfrac{y^5}5 +...\right)$
$\ln x=2\left(\left(\dfrac{x-1}
{x+1}\right)+\dfrac13{\left(\dfrac{x-1}{x+1}\right)^3}+\dfrac15{\left(\dfrac{x-1}{x+1}\right)^5}+...\right)$
If we measure the radius of convergence with respect to $y$, we get just one unit inherited from the original Maclaurin series. But with respect to $x$ we see that $-1<y<1$ and therefore this series converges to $\ln x$ for all positive $x$!
