Find the largest interval for which the following series is convergent at all points x in it Find the largest interval for which the following series is convergent at all points x in it.
\begin{equation} \sum_{n=1}^{\infty} \frac{2^n(3x-1)^n}{n}. \end{equation}
Applying ratio test radius of convergence is 2, then we get $(-\frac{1}{3}, 1)$ But the difficulty is at extreme points.Please help.
 A: We can apply the ratio test either to the original series, which is an expansion around $x=1/3$, or we can apply it to the transformed series by taking $z = 3x-1$.
If we do the former, the ratio test gives a radius of convergence of $1/6$ around $x=1/3$.  If we do the latter, the radius of convergence is $1/2$ around $z=0$.  So either $x \in (1/6,1/2)$ or $z \in (-1/2,+1/2)$ will be an interval of absolute convergence, according to the ratio test.
Now we get an alternating series at the endpoint $x=1/6$, equiv. $z=-1/2$, and because of the $n$ in the denominator, this will be a conditionally convergent series (harmonic) we are familiar with.  
At the other endpoint the series is all positive terms/harmonic, and we know that diverges (very, very slowly).
A: Convergence of $\begin{equation} \sum_{n=1}^{\infty} \frac{2^n(3x-1)^n}{n} \end{equation}$ would actually mean the requirement of 
$$|2(3x-1)|<1\Rightarrow \frac{1}{6}<x < \frac{1}{2}$$
So.. It is assured that that converges for $$\frac{1}{6}<x < \frac{1}{2}$$ but this may not be the largest.. 
so you have to check for end points... (Which would give you "the largest")
I would leave that to you....
A: Replace the numerator by $y^n$. What you then face is the infinite Taylor series of $-\log (1-y)$. Replace now by the definition you used for $y$ and you get $-\log (3 - 6 x)$. Then, ...
