# Cauchy Sequences

• October 23rd 2009, 02:59 AM
freakinbears
Cauchy Sequences
Hello everyone in the math community,

Here's a problem I hacked at for a bit and can't quite seem to find an adequate solution for (regardless of my approach, the differences I set up in absolute value never quite seem to cancel properly).

Suppose $x_n$ is a Cauchy sequence (the usual properties; for all epsilon, there is an index beyond which the distance between two elements is less than epsilon). In addition, $x_n$ has the property that for all epsilon, there exists an index k > 1 / epsilon, such that: $|x_k| <$ epsilon.

Prove that $x_n$ converges to 0.

Now I understand the gist of the image; arbitrarily close to 0 we will find SOME term of the sequence close to 0, and the Cauchy property will force the rest of the sequence to lie near 0. However, the actual mechanics of the proof aren't quite clicking for me. Help? :)
• October 23rd 2009, 04:48 AM
HallsofIvy
I don't see why you need "Cauchy" at all! If you have that "$x_n$has the property that for all epsilon, there exists an index $k > 1/\epsilon$, such that: $|x_k|< \epsilon$", that's all you need.

You want to show that "given $\epsilon> 0$, there exist N such that if k> N, then $|x_k- 0|= |x_k|< \epsilon$. Okay, just take $N= 1/\epsilon$.
• October 23rd 2009, 05:03 AM
freakinbears
But it's not that simple! The condition only guarantees a single term of the sequence to be within an arbitrary epsilon. It says, for any epsilon, I can return you a term of the sequence which is within epsilon of 0. However, it gives no clue as to the next term in the sequence, which could, for all we know, be 34. The only reason we have any inference into the general behavior of the sequence is that we are given it is Cauchy (hence preventing the '34' case).
• October 23rd 2009, 05:27 AM
freakinbears
Here's my attempt:

Let epsilon>0. Then exists N st. m,n>N implies |x_n-x_m| < epsilon.
Let delta>0, sufficiently small so that 1/delta > N. Then choose n_delta such that |x_n_delta| < delta. Because 1/delta > N, n_delta > N.

Hence |x_n| = |x_n - x_n_delta + x_n_delta| <= |x_n - x_n_delta| + |x_n_delta| < epsilon + delta.

If this answer were instead epsilon, the proof would be finished, and x_n would converge to 0. Some help with this?
• October 23rd 2009, 07:17 AM
rn443
Quote:

Originally Posted by freakinbears
Hello everyone in the math community,

Here's a problem I hacked at for a bit and can't quite seem to find an adequate solution for (regardless of my approach, the differences I set up in absolute value never quite seem to cancel properly).

Suppose $x_n$ is a Cauchy sequence (the usual properties; for all epsilon, there is an index beyond which the distance between two elements is less than epsilon). In addition, $x_n$ has the property that for all epsilon, there exists an index k > 1 / epsilon, such that: $|x_k| <$ epsilon.

Prove that $x_n$ converges to 0.

Now I understand the gist of the image; arbitrarily close to 0 we will find SOME term of the sequence close to 0, and the Cauchy property will force the rest of the sequence to lie near 0. However, the actual mechanics of the proof aren't quite clicking for me. Help? :)

Pick any epsilon. Then there exists an N such that for all n, m >= N |x_n - x_m| < epsilon/2. Thus, we need only show that there is an x_k with |x_k| < epsilon/2 and k >= N (since in that case for all greater j, |x_j| <= |x_j - x_k| + |x_k - 0| < epsilon). But this is guaranteed by the fact that there's a subsequence that converges to zero.
• October 23rd 2009, 07:21 AM
HallsofIvy