Infinite Series - Ratio Test (1 Viewer)

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Hi all!
Here's something I'm having difficulty seeing:
Suppose

[tex]u_n > 0[/tex] and

[tex]\frac{u_{n+1}}{u_n} \leq 1-\frac{2}{n} + \frac{1}{n^2}[/tex] if [tex]n \geq 2[/tex]

Show that [tex]\sum{u_n}[/tex] is convergent.

I'm not sure how to apply the ratio test to this.
It looks like I would just take the limit.

I get: [tex]lim_{n \rightarrow \infty} 1-\frac{2}{n} + \frac{1}{n^2} = 1[/tex]

I'm not sure if I'm correct, but I could see this two ways.
Since the above limit converges to 1, then the summation converges by the ratio test.
Or, since the limit converges to one, the summation may converge or diverge.
Is either statement correct? Am I on the right track?
Thanks for the help.
Bailey
 
Got it...

[tex]u_n=\frac{k}{(n-1)^2}[/tex] where k is a constant.

This is just the series:

[tex]k \sum{\frac{1}{n^2}}[/tex] which we know converges


Whew!

Bailey
 

benorin

Homework Helper
1,057
7
Couldn't there exist other series which satisfy said inequality?

Try Gauss' convergence test for series.
 
Since the ratio test demands that the limit be less than 1, it looks as though the ratio test fails as Mr. Bailey have shown.
 
694
0
Couldn't there exist other series which satisfy said inequality?
He probably meant u_n <= k/(n - 1)^2.

Let a_n = 1 - 2/n + 1/n^2.

Then u_(n + 1) <= a_n * u_n <= a_n * a_(n - 1) * u_(n - 1), etc. Inductively, we have that u_(n + 1) <= a_n * a_(n - 1) * ... * a_1 * u_0.

But as "luck" would have it, a_n * a_(n - 1) * ... * a_1 = 1/(n - 1)^2 (easy to show with induction), so the desired inequality follows. (N.B the details are probably not all correct. But that's relatively unimportant).
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top