Cauchy Sequence: Understanding the Boundary Condition

[steven187]

hello all

I found this rather interesting
suppose that a sequence {x_{n}} satisfies

|x_{n+1}-x_{n}|<\frac{1}{n+1} \forall n\epsilon N

how couldn't the sequence {x_{n}} not be cauchy? I tried to think of some examples to disprove it but i didnt achieve anything doing that, please help

thanxs

 

[HallsofIvy]

An obvious one: just add those differences. Let x_n be \sum_{i=1}^n \frac{1}{n}. That series does not converge and so the sequence of partial sums is not Cauchy.

 

[saltydog]

I'm hesitant to challenge you Hall but feel compelled to do so at grave risk as Analysis is not my strong suite. Alright nothing is but I digress.

It looks like a Cauchy sequence to me based on the definition:

The sequence: \{x_n}\} is a Cauchy sequence if given \epsilon there exists N such that for all m,n\leqN we have:

|x_m-x_n|<\epsilon

For the sequence:

|x_{n+1}-x_{n}|<\frac{1}{n+1}


I take \epsilon=\frac{1}{N+1};

Thus for all n>N:

|x_{n+1}-x_n|<\frac{1}{n+1}<\frac{1}{N+1}=\epsilon

 

[rachmaninoff]

Thus for all n>N:
|x_{n+1}-x_n|<\frac{1}{n+1}<\frac{1}{N+1}=\epsilon

That's true; but look closely - is that really the same thing as

for all m,n\geqN we have:<br /> |x_m-x_n|&lt;\epsilon

?

 

[HallsofIvy]

For a given &epsilon;, you then choose a large enough N. No fair choosing N first and then finding &epsilon; so that N doesn't work!
You do remind me that just |x_{n+1}-x_n|&lt; \epsilon is not enough. You have to show that |x_n-x_m|&lt; \epsilon which is basically showing that the tail of the sum goes to 0.

 

[saltydog]

That's true; but look closely - is that really the same thing as



?

I suppose not.

|x_{n+1}-x_n|&lt;\frac{1}{n+1}

does not imply

|x_m-x_n|&lt;\frac{1}{n+1}

Still a little unclear. I'll work on it. Thanks guys.

Seriously, why do I even bother with differential equations anyway . . .

 

[rachmaninoff]

Simply put: for given N, your

| x_m - x_n |,

\mbox{ as } m \rightarrow \infty,

is not guaranteed to be any better than the (divergent) harmonic series:

| x_m - x_n | &lt; \sum_{k=n+1}^{\infty} \frac{1}{n} \rightarrow \infty

 

[saltydog]

Alright, I'm interested in getting my math right.
Following Hall's example (a slight change):

Let:

x_n=\sum_{i=1}^{n+1} \frac{1}{i}

thus:

|x_{n+1}-x_n|=|\sum_{n=1}^{n+2} \frac{1}{i} - \sum_{n=1}^{n+1} \frac{1}{i}|=\frac{1}{n+2}

thus:

|x_{n+1}-x_n|=\frac{1}{n+2}&lt;\frac{1}{n+1}

However, the partial sums of this harmonic series \{x_n\} do not converge and thus cannot be a Cauchy sequence.

Now, I got a pot of spagetti to make . . .

 

[steven187]

hello all

nice stuff saltydog, i had a similar result, i think your fine with analysis, and hallsofivy your right it is an obvious one but the funny thing is i can't think of any other examples, do you know of any others?

steven

 

[Kevin"12]

hi guys i need bounded but not cauchy function can you help me asap?

 

[alexfloo]

Kevin, consider any bounded periodic function that is not constant (like, for instance, a sinusoid). It will never become and remain arbitrarily close to anything, but it does remain within fixed bounds.