Sequences limits and cauchy sequences

In summary, for every m,n big enough a(m)-a(n)<epsilon, so a(n) is a cauchy sequence. If you take m to be 2^n or something like that, then a(n) is a cauchy sequence. For n<m, a(n)-a(m)=(a(n)-a(n+1))+(a(n+1)-a(n+2))+(a(n+2)-a(n+3))+ ... (a(m-1)-a(m)).
  • #1
ibc
82
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Homework Statement


prove or refute:

if lim(a(2n)-a(n)=o , then a(n) is a cauchy sequence


Homework Equations





The Attempt at a Solution


I need to prove that for every m,n big enough a(m)-a(n)<epsilon
so I know for all m and n I can say m=l*n, lim(a(m)-a(n))=lim(a(n*l)-a(n*l/2) +a(n*l/2) -a(n*l/4)...+a(2n)-a(n)), which is the sum of a lot of zeros, though if I take m to be 2^n or something like that, I get an inifinite amount of zeros, so I don't know what I can do with that.
so I tried to find a sequence which contredicts it, though couldn't find any




Homework Statement


prove or refute:
if |a(n+1)-a(n)|<9/10*|a(n)-a(n-1)|
then a(n) is a cauchy sequence


Homework Equations






The Attempt at a Solution


well fromt he equation I can get:
(|a(n+1)-a(n)|)/(|a(n)-a(n-1))<9/10<1
so what it gives me is that all that in the power of n is going to zero, which means is a cauchy sequence, thoguh I don't see how it helps me =\
 
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  • #2
Yes, you want a counterexample for the first one. It's going to have to be a function that grows REALLY slowly. What kind of functions can you think of that do that? For the second one for n<m, a(n)-a(m)=(a(n)-a(n+1))+(a(n+1)-a(n+2))+(a(n+2)-a(n+3))+ ... (a(m-1)-a(m)). Think you can maybe bound that sum by a geometric series?
 
  • #3
Dick said:
Yes, you want a counterexample for the first one. It's going to have to be a function that grows REALLY slowly. What kind of functions can you think of that do that? For the second one for n<m, a(n)-a(m)=(a(n)-a(n+1))+(a(n+1)-a(n+2))+(a(n+2)-a(n+3))+ ... (a(m-1)-a(m)). Think you can maybe bound that sum by a geometric series?

I know I need something that grows really really slow, but I couldn't find any
and we can't use functions yet in the course, so it'd have to be a sequence/series
the slowest one I could think of is 1/1+1/2+1/3+1/4+... which is good to refute the another question, which was "if lim(a(n+1)-a(n)=o , then a(n) is a cauchy sequence", though for this one it's not helpful.
 
  • #4
a(n)=log(n) grows pretty slowly, but not quite slowly enough. Try sqrt(log(n)).
 

1. What is a sequence limit?

A sequence limit is the value that a sequence approaches as the index approaches infinity. In other words, as the terms in a sequence get closer and closer to a specific number, that number is considered the limit of the sequence.

2. How do you find the limit of a sequence?

To find the limit of a sequence, you can use the following formula: lim(n→∞) an = L, where L is the limit and an is the nth term in the sequence. Alternatively, you can also graph the sequence and observe the trend of the terms as the index increases.

3. What is a Cauchy sequence?

A Cauchy sequence is a sequence in which the terms get closer and closer together as the index increases. In other words, for any positive number ε, there exists a point in the sequence after which all terms are within ε distance of each other.

4. How do you determine if a sequence is a Cauchy sequence?

To determine if a sequence is a Cauchy sequence, you can use the Cauchy convergence criterion, which states that a sequence is Cauchy if and only if for any positive number ε, there exists a point in the sequence after which all terms are within ε distance of each other.

5. What is the difference between a sequence limit and a Cauchy sequence?

The main difference between a sequence limit and a Cauchy sequence is that a sequence limit is the value that a sequence approaches, while a Cauchy sequence is a sequence in which the terms get closer and closer together. A sequence can have a limit without being a Cauchy sequence, and vice versa.

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