Infinite Series with log natural Question

Biosaw
Messages
1
Reaction score
0

Homework Statement



Ʃ (-1)^(k+1) / kln(k)
k=2

Homework Equations


integral test, p test, comparison test, limit comparison, ratio test, root test.


The Attempt at a Solution


In class so far we have not learned the alternating series test so i can't use that test.
So far I have done the test for divergent which turned up 0, ratio test which turned up 1, root test which didn't work, and its not a geometric or telescoping series.
I think that the only one that will work is the comparison test but I don't know what to use to compare this function.
 
Physics news on Phys.org
From your question I gather that you only need to say whether the series converges, not what the limit is.

Never mind my old post, it doesn't seem to help.

[old post]
I am not sure how much you have and haven't seen, but one common approach would be looking at the absolute value and using the triangle inequality to write

\left| \sum_{k = 2}^\infty \frac{(-1)^{k + 1}}{k \ln k} \right| \le \sum_{k = 2}^\infty \left| \frac{(-1)^{k + 1}}{k \ln k} \right|

It may also be useful to note that ln k > 1 for k > 2.
[/old post]
[/color]

[new hint]
Whichever way you try the comparison test, it will always end up "going the wrong way".
Try an integral test instead.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top