# Progression, is monotonic?

1. Oct 4, 2008

### Physicsissuef

1. The problem statement, all variables and given/known data

I got one progression $$\frac{2n-3}{3n-5}$$.

Is this monotonic and convergent?

2. Relevant equations

3. The attempt at a solution

I tried an-an+1=1/(3n-5)(3n-2)>0

But for n=1 and n=2, we got 1/2 and 1, so I think that this string is not monotonic, right?

I think it is convergent because it got a=2/3.

$$|a_n-a|<\epsilon$$

Am I right?

2. Oct 4, 2008

Are you sure

$$a_n - a_{n+1} = \frac 1 {(3n-5)(3n-2)}$$

is $$> 0$$ for every $$n = 1, 2, 3, \dots$$? Compare the result you
get from the formula above to the actual value of $$a_1 - a_2$$.
You are on the correct track for proving convergence.

3. Oct 4, 2008

### tiny-tim

Hi Physicsissuef!

It would be much easier if you you rewrote it in the form 2/3 + A/(3n - 5).

If it's monotonic except for the first one or two, then you should say so.

It converges monotonically after a certain point.

4. Oct 5, 2008

### stmartin

Yes, I am sure that it is correct, but as you can see for n=1 (i.e a1,a2), it is

1/(3-5)(3-2)
-1/2 < 0

and for n>1, it is monotonic.

So if I write
$$|a_n-a|=|\frac{2n-3}{3n-5}-\frac{2}{3}|=|\frac{3(2n-3)-2(3n-5)}{3(3n-5)}|= |\frac{6n-9-6n+10}{3(3n-5}|=\frac{1}{3(3n-5)} < \epsilon$$

So it is convergent, probably for n>1, because if n=1, $$1/3(3-5)=-1/6 < \epsilon$$

Because of the fact that $$\epsilon$$ can't be negative i.e $$\epsilon > 0$$, it is convergent for n>1, right?