Does the Ratio Test guarantee convergence for this infinite series?

[MrBailey]

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

u_n > 0 and

\frac{u_{n+1}}{u_n} \leq 1-\frac{2}{n} + \frac{1}{n^2} if n \geq 2

Show that \sum{u_n} 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: lim_{n \rightarrow \infty} 1-\frac{2}{n} + \frac{1}{n^2} = 1

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

 

[MrBailey]

Got it...

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

This is just the series:

k \sum{\frac{1}{n^2}} which we know convergesWhew!

Bailey

 

[benorin]

Couldn't there exist other series which satisfy said inequality?

Try Gauss' convergence test for series.

 

[robert Ihnot]

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.

 

[Muzza]

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).