# Thread: Absolute or conditional convergence

1. ## Absolute or conditional convergence

Ok, so I know that this series converges, but can someone please show me how I would determine absolute or conditional? Thanks!

$\displaystyle \sum^{\infty}_{n=3} (-1)^{n-1} \frac{ln(n)}{n}$

2. Does the sequence $\displaystyle \frac{\ln (n)}{n}$ decrease and converge to zero?
That is the alternating series test.

Now does the series $\displaystyle \sum\limits_{n = 3}^\infty {\frac{{\ln (n)}}{n}}$ converge?
If not then the convergence is not absolute.

3. Originally Posted by Plato
Does the sequence $\displaystyle \frac{\ln (n)}{n}$ decrease and converge to zero?
That is the alternating series test.

Now does the series $\displaystyle \sum\limits_{n = 3}^\infty {\frac{{\ln (n)}}{n}}$ converge?
If not then the convergence is not absolute.
I understand all that. What I am trying to figure out is since I have shown that $\displaystyle \sum\limits_{n = 3}^\infty {\frac{{\ln (n)}}{n}}$ does indeed converge. Does it converge conditionally because of the $\displaystyle (-1)^{n-1}$? This was a question on a test that I had this morning. It was a two part question. The first part I had to show what test I would use to figure out if it converged or not. I used the Alternating Series test and found that $\displaystyle b_n = 0$, $\displaystyle a_{n+1} < a_n$, and that it was alternating, which proves that the series converges. For part b, it asked me to show if it converged conditionally or absolutely. I said it converged conditionally. It converged because it met the criteria for an alternating series. However, with$\displaystyle (-1)^{n-1}$ I said that by having this as part of the series, it would affect its convergence because it caused the values to bounce around and not hone in on a specific value. Is that right or no?

Thanks for the response!

4. Originally Posted by mollymcf2009
I understand all that. What I am trying to figure out is since
I have shown that $\displaystyle \color{red}\sum\limits_{n = 3}^\infty {\frac{{\ln (n)}}{n}}$ does indeed converge.
How could you understand all this if you think you have proved the above in red.
The fact is: $\displaystyle \sum\limits_{k = 3}^\infty {\frac{{\ln (k)}}{k}}$ diverges.

It should have been worked out in the textbook.

Just note that $\displaystyle k \geqslant 3\, \Rightarrow \,\frac{{\ln (k)}}{k} > \frac{{k - 1}}{{k^2 }}$.

5. Originally Posted by Plato
How could you understand all this if you think you have proved the above in red.
The fact is: $\displaystyle \sum\limits_{k = 3}^\infty {\frac{{\ln (k)}}{k}}$ diverges.

It should have been worked out in the textbook.

Just note that $\displaystyle k \geqslant 3\, \Rightarrow \,\frac{{\ln (k)}}{k} > \frac{{k - 1}}{{k^2 }}$.
Wow, ok, sorry. I guess I may have left something out of the question, it is just from memory. I do understand this concept, I just wanted some clarification on the process of conditionally vs. absolute. I'll see if I can clarify the actual problem with my professor and re post.

Ok, so I got a response back from my professor. It is true that the series $\displaystyle \frac{ln(n)}{n}$diverges. But with the $\displaystyle (-1)^{n-1}$it converges conditionally. I was confused, because I just thought that if a series is divergent, it's divergent. Period. I think I just assumed that for a series to be conditionally convergent, it began as convergent and due to making a change to it i.e. $\displaystyle (-1)^{n-1}$ made it be conditionally convergent. Anyway, looks like I solved it correctly on my test, but didn't look at the series long enough to realize that it was actually divergent.