# Thread: convergence of piecewise function

1. ## convergence of piecewise function

Hello,

I have the following sequence:

$a_{n}=\frac{1}{\ln(x)}$ when n is even
$a_{n}=-\frac{1}{\ln(x)}$ when n is uneven

I want to determine whether the sequence is absolute convergent, divergent or conditional convergence.

I have tried showing that the sequence is absolute covergent by showing that $\sum_{2}^{\infty} \vert a_{n} \vert$ is convergent. I tried using the integral test but I ended up with a strange expression (which contains li(x)).
I have also tried using the ration test which leads to no conclusion.

Suggestions would be appreciated.

Thanks

2. ## Re: convergence of piecewise function

I presume you mean "log(n)", not "log(x)"? The problem is that this is not absolutely convergent.

(If you do mean " $\frac{1}{log(x)}$ for constant "x" then if it very easy!)

3. ## Re: convergence of piecewise function

Sorry but I presented the problem in a wrong way.

I have the sequence $a_{n}$ given in the previous post.

How do I determine if the series $\sum_{2}^{\infty} a_{n}$ is absolutely convergent, conditionally convergent or divergent.

I say that the series is divergent since $\lim_{n \rightarrow \infty} a_{n} \neq 0$.

What say you?

4. ## Re: convergence of piecewise function

$\displaystyle \lim_{n \to \infty} \dfrac{1}{\log{n}} = 0$

$\displaystyle \sum_{n=2}^\infty \dfrac{(-1)^n}{\log{n}}$ converges by the alternating series test

$\displaystyle \sum_{n=2}^\infty \dfrac{1}{\log{n}}$ diverges by direct comparison with $\displaystyle \sum_{n=2}^\infty \dfrac{1}{n}$, a known divergent series.