# Proove series question

1. Nov 30, 2008

### transgalactic

http://img372.imageshack.us/img372/7929/76106583tn6.gif [Broken]

??

Last edited by a moderator: May 3, 2017
2. Nov 30, 2008

### Office_Shredder

Staff Emeritus
This looks like a question about sequences, not series. Have you tried doing anything yet? The definition of convergence seems like a good place to start

3. Nov 30, 2008

### transgalactic

this question is obvious
its common sense

if a(n) converges to a

then if we take the function who takes the biggest member
of course it will pick the closest to "a"

i dont know how to transform these word into math

??

4. Nov 30, 2008

### Office_Shredder

Staff Emeritus
Definition of convergence: For all e>0, there exists N>0 such that n>N implies |an - a| < e

So.... if you take n>N, what can you say about |bn-a|?

5. Nov 30, 2008

### transgalactic

|bn-a|<e

this inequality shows that b(n) has not reached the bound

what is the next step?

6. Nov 30, 2008

### Office_Shredder

Staff Emeritus
So given e>0, can you find N such that n>N implies |bn-a| < e?

7. Nov 30, 2008

### transgalactic

whats the role of "N"

why are you writing N>0 n>N ??

instead you could write just n>0

??

8. Dec 1, 2008

### VeeEight

It is not for any n > 0 . For example, consider the sequence {1, 1, 3, 3, 4, 5, 5, 5, 5, 5, 5, 5, 5 ..... etc} that converges to 5. Here our N is 6 since for any n > N, |a_n - 5| < e. So the role of N is to show that once you pass a_N, the sequence has essentially converged.

Think about what is going on in the sequence b. Are it's entries similar to those in a_n? What happens when you take a very large n?

9. Dec 1, 2008

### transgalactic

why in this example our N is 6?

the definition of a couchy sequesnse is:
after a finite number of steps, any pair chosen from the remaining terms will be within distance "e" of each other.

i can guess that for each bn

it take the largest member from a1 to an

for example:
a1=6 a2=5 a3=6 a4=1 a5=8 a6=7

b1=6 b2=6 b3=6 b4=6 b5=8 b6=8

what to do now?

Last edited: Dec 1, 2008
10. Dec 1, 2008

### transgalactic

anyone?

11. Dec 1, 2008

### VeeEight

N is 6 because after 6 steps, any pair of terms is "0" away from each other, so they are certainly less than "e" away from each other. So in general, N is the number of finite steps you must advance before "any pair chosen from the remaining terms will be within distance "e" of each other". This is more of an aid to visualize things and to keep things formal.

You may want to consult some of the theorems from real analysis like monotone convergence.

12. Dec 1, 2008

### Office_Shredder

Staff Emeritus
We're not talking about Cauchy sequences, we're talking about convergent sequences. They happen to be equivalent, but when you're proving something is convergent it's often easier to just prove it's convergent. I posted the definition of a convergent sequence above. In the example Vee gave, N (in the definition of convergence, whose relevance I'll go into deeper later) is 6 since after 6 steps, you reach the limit point (and hence for any e>0, |an - 5| = 0 < e.

The point of saying N>0 isn't to indicate that n>0 later, the point is that you have to pick a very large number (like you said, some finite number of steps). So given e>0, there exists N (whose value you don't know) such that n>N implies |an - a|<e. For small e, N is going to be something like 10,000 or 100,000. This isn't the same as just saying n>0

13. Dec 1, 2008

### transgalactic

ok i understand after N step we have the members |bN-b(N+1)| < e
and when we take n>N its an over kill(to be absolutely sure)

Last edited: Dec 1, 2008
14. Dec 1, 2008

### transgalactic

but in this question i dont have the formula of sequence "a"

so i cant do this process

how to apply this process on my question?

15. Dec 1, 2008

### transgalactic

i am only given that sequence An converges to the value "a"

16. Dec 3, 2008

### transgalactic

i could say that if An s monotonic and converges then An+1 also converges to "a"

so Bn ->a
but here i cant say that An is monotonic

??

Last edited: Dec 3, 2008