For what values of $x$ is the series $\sum_{k=2}^{\infty}\frac{(x-2)^k}{k\ln{k}}$ convergent? For what values of $x$ is the series $$\sum_{k=2}^{\infty}\frac{(x-2)^k}{k\ln{k}}$$ absolutely convergent, conditionally convergent and divergent?
Denote $a_k=(x-2)^k/k\ln{k},$ I get
$$\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_k}\right|=|x-2|\lim_{k\rightarrow\infty}\frac{k\ln{k}}{(k+1)\ln{k+1}}=1,$$
according to L'Hopitals rule. So I have that $-1 \leq x-2\leq1$ which means that the series is absolutely converging for $1<x<3$. Checking the endpoints:
$x=1:\Rightarrow$
$$\sum_{k=2}^{\infty}\frac{(-1)^k}{k\ln{k}}.$$
This is an alternating series. Denote $b_k=\frac{1}{k\ln{k}}.$ I have to show that $\lim_{k\rightarrow\infty}b_k=0$ and that $b_k$ is a decreasing function. The first condition is trivial. The second condition is trivial to since it is easy to see that $b_k>b_{k+1}>b_{k+2}>...>b_{k+m},$ for $m\in\mathbb{N}.$ So this is a conditional convergence.
$x=3:\Rightarrow$
$$\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$$ and the integraltest gives
$$\int_{x=2}^{\infty}\frac{1}{x\ln{x}}=\lim_{x\rightarrow \infty}\log(\log{x})=\infty$$
which diverges and the series diverges too for $x=3$. Thus the series behaviour can be summed up as follows:
The radius of convergence is $R=1$ and the interval of convergence is $I=[1,3).$ The series is


*

*Absolutely convergent for $x\in(1,3)$

*Conditionally convergent for $x=1$

*Divergent for the rest.


Is this correct?
EDIT: What I don't understand is that in my book the following theorem about convergence is stated:

  
*
  
*If the series $\sum|a_k|$ converges, then the series $\sum a_k$ also converges. This series is then absolutely convergent.
  
*If the series $\sum |a_k|$ is divergent and $\sum a_k$ is convergent, then the series $\sum a_k$ is conditionally convergent.
  

But in my solution above, I don't see anywhere this theorem is applied. I've just gonv for the limit of $|a_{k+1}/a_k|$, not the sum of $|a_k|$. Can someone explain this?
 A: Using ratio test, one claims that for all $x$ with $|x-2|<1$, the series converges absolutely. For $x-2=1$, the series is $\displaystyle\sum_{k=2}\dfrac{1}{k\log k}$ which does not converge. For $x-2=-1$, the series is $\displaystyle\sum_{k=2}\dfrac{(-1)^{k}}{k\log k}$ which is convergent by alternating series test, but not absolutely, so it is conditional.
A: Hints:


*

*The Bertrand's  series $\displaystyle\sum_{k\ge2}\frac1{k\,\ln k}$ is divergent.

*If $a>1$, $k\ln k=o(a^k)\strut$.

A: You should say that it's "conditionally convergent for $x=1$" and "absolutely convergent for $x\in \color{red}{(}1,3)$", and I don't believe your argument that $\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$ diverges is valid.  
The comparison that $\frac{1}{k \ln k} < \frac{1}{k}$ would have teeth if $\sum_{k=2}^{\infty}\frac{1}{k}$ converges, and it would show that $\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$ also converges.  As we know it doesn't.  If 
$\frac{1}{k \ln k} > \frac{1}{k}$ were somehow true, then you could argue that since $\sum_{k=2}^{\infty}\frac{1}{k}$ diverges $\sum_{k=2}^{\infty}\frac{1}{k\ln{k}}$ definitely diverges, but we know that's not true either.    
The integral test is the most basic way to show it diverges I think.
$\bigg [\int \frac{1}{x \ln x} dx=\int \frac{1/x}{\ln x} dx \left(\to \int \frac{du}{u}\right)=\ln(\ln x) \to \infty \ \ \text{as} \ \ x \to \infty \bigg] \implies \bigg [\sum_{k=2}^{\infty}\frac{1}{k\ln{k}} \to \infty \bigg]$
Accordingly, as user284331 pointed out, you should say that $\sum_{k=2}^{\infty}\frac{(x-2)^k}{k\ln{k}}$ converges conditionally at $x=1$
