Convergence of a Series: How to Determine its Value?

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 2K views
MeMoses
Messages
127
Reaction score
0

Homework Statement


I'm not sure how to do the notation on here but. Does this series converge or diverge. If it converges, then to what value.
The series: Sum from 1 to infinity of [(-1)^n * n / (n^2-4n-4)]


Homework Equations


It tells me to use the ratio test


The Attempt at a Solution


I used the ration test and got the limit to equal 1 which is inconclusive. I used the alternating series test to find that it converges, but how do I figure what it converges to and whether it is absolute convergence or not? I'm new to series so this is a little confusing. Any help is appreciated
 
Physics news on Phys.org
In LaTeX form, this is what your equation appears like:
[tex]\sum^{\infty}_{n=1} \frac{(-1)^n n}{ (n^2-4n-4)}[/tex]
Since there is an alternating sign and you need to find the series, i believe you need to use AST.

The conditions that must be satisfied for the series to converge:[tex]a_n>0<br /> \\\lim_{n \rightarrow \infty} a_n=0<br /> \\a_{n+1} \leq a_n[/tex]where[tex]a_n=\frac{n}{n^2-4n-4}[/tex]Indeed, the series converges.
[tex]\lim_{n \rightarrow {\infty}} \frac{n}{n^2-4n-4}=\lim_{n \rightarrow {\infty}} \frac{1/n}{1-4/n-4/n^2}[/tex]
I would say it converges to 0?

To test for absolute or conditional convergence, test if the absolute value of the original series converges or not:
[tex]\sum^{\infty}_{n=1} \left|\frac{n}{ (n^2-4n-4)}\right|[/tex]
 
Last edited:
How can you say it converges at 0? When the limit of An is 0 doesn't that just mean that the series converges? Wouldn't I need to write it as a function somehow and take the limit of that to find the convergence? This is an online problem and 0 isn't correct
 
Sorry I was reading the questing wrong. It was asking for my results from the ratio test which explains a lot.
 
MeMoses said:
How can you say it converges at 0? When the limit of An is 0 doesn't that just mean that the series converges?
No, it means the series may converge. It may still diverge as well. In any case, what sharks implied above is wrong, as you noted.

Wouldn't I need to write it as a function somehow and take the limit of that to find the convergence? This is an online problem and 0 isn't correct
Yes, generally you'll need to find another way to figure out what a series converges to, if it indeed converges.