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Show that if $ \{ p_n \} $ is a Cauchy sequence, then it has a subsequence $ \{ p_{n_k}\} $ such that the series $ \sum_{k=1}^\infty b_k $ converges, where $ b_k = d(p_{n_k}, \, p_{n_{k+1}}) $.

My approach

we know that if $p_n$ is cauchy sequence, it has some subsequence that converges. That means that the $d(p_{n_k}, \, p_{n_{k+1}})\to 0$ for big $k$.

Hence the sum of $b_k$ also converges. I think it is right but can someone put it in more formal complete steps? I think I am missing few steps and I am not sure how to write it more formally than this.

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  • $\begingroup$ What space are we working with? $\endgroup$ – Mathmo123 Jun 23 '14 at 0:11
  • $\begingroup$ your approach is not fully correct. you need to assure the distances are small enough, say assure that $b_k \leq 2^{-k}$ or any other series that you know to be convergent $\endgroup$ – mm-aops Jun 23 '14 at 0:14
  • $\begingroup$ Its not enough for $d(p_{n_k},p_{n_{k+1}})\to 0$. But you can make $d(p_{n_k},p_{n_{k+1}})$ go to zero as fast as you want... fast enough to make the series converge. $\endgroup$ – Zarrax Jun 23 '14 at 0:14
  • $\begingroup$ This is very close to being true by definition. Since the given sequence is Cauchy, you can choose distances between successive terms to be as small as you like. So the statement is hardly saying anything more than "there exists a convergent infinite series of positive terms". (OK I guess it requires the comparison test too.) $\endgroup$ – David Jun 23 '14 at 0:23
  • $\begingroup$ What are $p_n$? What is $d$? $\endgroup$ – tomasz Jun 23 '14 at 1:46
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Note that your assertion that $p_n$ has a convergent subsequence is only true if we're working in a complete metric space.

Since $p_n$ is Cauchy, that means that given $k \in \mathbb N, \exists N_k$ such that $\forall n,m\ge N_k,$ $d(p_n, p_m) < \frac{1}{2^k}$

Now consider the subsequence $p_{N_k}$. Can we use this in any way?

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  • $\begingroup$ One small nitpick : You need to ensure that $N_{k+1} \geq N_k$, which means you need to proceed by induction. $\endgroup$ – Prahlad Vaidyanathan Jun 23 '14 at 4:12
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Just because $b_k = d(p_{n_k},p_{n_{k+1}}) \to 0$ does not mean that $\sum b_k$ converges.

But, we can use this to find a convergent subequence.

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Chose $p_{n_k}$ inductively. Given $n_k$, let $N$ be such that if $n,m \geq N$ then $|p_n-p_m|< \frac{1}{2^k}$. Now chose $n_{k+1}=N$ This will insure that $$b_{k+1}=d(p_{n_{k+1}},p_{n_{k+2}}) <\frac{1}{2^k}.$$ (regardless of how $n_{k+2}$ is chosen at the next step.)

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