I'm trying to do this by induction, I can actually show this for the case k=n. I cannot however do it for the case =1.

Also

Can anyone help me show>?

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- Sep 23rd 2008, 07:06 PMhockey777Show a sequence is decreasing

I'm trying to do this by induction, I can actually show this for the case k=n. I cannot however do it for the case =1.

Also

Can anyone help me show__>__? - Sep 24th 2008, 03:41 AMhockey777
Sorry, I was editing while you were responding.

- Sep 24th 2008, 07:17 AMhockey777
Actually I just need to show it's

Monotone - Sep 24th 2008, 08:10 AMbkarpuzSome arguments
**Section A**

Let be**nonincreasing**and .

Then, we have

which indicates by induction that for all .

Clearly, exists and holds.

Passing limit as on both sides of the recursion relation, we get or equivalently .

Thus, must be true.

However, I really do not think that the given relation and the intitial condition gives .

**Section B**

But if you can show , the rest follows by induction since

is true if holds.

I will be thinking on it...

**Section C**

Set for .

Hence, it is easy to see that for all since holds for all .

Now, we shall show , which implies .

Apply*Section B*to obtain the result.

**Section D**

Therefore, is a decreasing sequence which tends to from above at . (Wink) - Sep 24th 2008, 08:30 AMhockey777
^To complicated, but got me on the right track

- Sep 24th 2008, 10:29 AMhockey777
Could someone help me show that 1 is the infinum of this set?

- Sep 24th 2008, 10:32 AMbkarpuz
- Sep 24th 2008, 11:25 AMhockey777
I saw that, but it's not really the approach I'm trying to master right now.

Does anyone else have anything else they can recommend? - Sep 24th 2008, 11:27 AMbkarpuz
- Sep 24th 2008, 11:32 AMhockey777
The question asks to prove it's monotone and convergent, then to find the limit.

I've proven it's monotone and convergent; by the Monotone Convergence Theorem I can show that the lim= inf( ).

What you did in your original post isn't tools we've used yet to prove infinums of sets yet. Hence, I've got to figure out another way to do it, and every way I"ve tried has ended in failure. - Sep 24th 2008, 11:48 AMbkarpuz
First, we show that its decreasing, next we show that it is bounded from below (in worst case it is bounded below by ), this means that it has a limit at infinity.

Then in the recursion we pass to limits on both sides (note that if then as ; that, is every subsequence converges to the same limit).

Finally, we prove that , which is the desired value since is decreasing.

I know I repeat the same, but no other idea to show . - Sep 24th 2008, 12:36 PMPlato
- Sep 24th 2008, 12:40 PMbkarpuz
- Sep 24th 2008, 12:59 PMhockey777
i figured it out and got it, thanks guys