Yes, that is correct. The sequence is convergent for n>1, but not for n=1.

[Physicsissuef]

Homework Statement

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

Is this monotonic and convergent?

Homework Equations

The Attempt at a Solution

I tried a_n-a_n+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?

 

[statdad]

Are you sure

<br /> a_n - a_{n+1} = \frac 1 {(3n-5)(3n-2)}<br />


is &gt; 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.

 

[tiny-tim]

Homework Statement

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

Is this monotonic and convergent?
…
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.

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.

 

[stmartin]

Are you sure

<br /> a_n - a_{n+1} = \frac 1 {(3n-5)(3n-2)}<br /> is &gt; 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.

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.

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.

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)}|=<br /> |\frac{6n-9-6n+10}{3(3n-5}|=\frac{1}{3(3n-5)} &lt; \epsilon

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

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