# 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.

• I think you have the wrong radius of convergence (or maybe just misstated it), but this is the right approach. Do the ratio test again, and don't forget the possibility of conditional convergence at the endpoints (limit imposed by radius of convergence). Jan 25, 2014 at 14:13
• sorry i made a transformation z=3x-1 in the power series. Jan 25, 2014 at 14:15
• Yes, that works. Now consider the endpoints, perhaps in terms of $z$ if it makes the computation easier. Jan 25, 2014 at 14:16
• yes. Is the series $2^{2n} /n$ convergent? Jan 25, 2014 at 14:17
• I'm pretty sure the proper interval is $(1/6, 1/2)$; not, $(-1/3,1)$. You need to check the endpoints, $x=1/2$ and $x=1/6$. Jan 25, 2014 at 14:23

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).

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....

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, ...