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I would like to construct a subsequence of the sequence $\large a_n = \frac{1}{n}$, so that the indices of the subsequence are determined by:

$n_1 = 1$ and $n_{k+1}$ is the smallest integer greater than $n_k$, such that if $m,n > n_{k+1}$, we have $|a_m - a_n| < 2^{-k-2}$.

Giova62 Gds
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  • What have you found for $n_k$? – Henry Jul 17 '23 at 19:30
  • Given $n=n_k+1,$ $$\lim_{m\to\infty}\left|\frac1n-\frac1m\right|=\frac1{n}=\frac1{n_k+1}.$$ Since this limit approaches strictly from below (all terms are strictly less than the limit,) you can take $n_k=2^{k+2}-1.$ Any smaller $n_k$ gives $n=n_k+1$ and some $m$ where the difference is too great. – Thomas Andrews Jul 17 '23 at 19:35
  • If $k=1$ then $n_k = n_1 = 1$. If $k=2$ then $n_k= n_2$.

    $n_2 > n_1$ and such that $|a_m - a_n| < 2^{-3}$

     – Giova62 Gds Jul 17 '23 at 19:48

2 Answers2

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If $m,n>n_{k+1}$ then $$\left|\frac1n-\frac1m\right|<\frac1{n_{k+1}+1}.$$

On the other hand, $n=n_{k+1}+1,$ then: $$\lim_{m\to\infty}\left|\frac1n-\frac1m\right|=\frac1{n}=\frac1{n_{k+1}+1}$$

So you want $\frac1{n_{k+1}+1}\leq \frac1{2^{k+2}},$ or $n_{k+1}\geq 2^{k+2}-1.$ So the smallest we can choose is $n_{k+1}=2^{k+2}-1.$

Thomas Andrews
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Can you explain the second part of the proof to me? An example of I1, I2, I3 with $a_n = 1/n$.

Theorem: A sequence of real numbers $(a_n)$ is convergent if and only if it satisfies the Cauchy condition.

Proof: The condition is necessary. If $\lambda$ is the limit of the sequence, then for every $\epsilon > 0$, there exists a $\overline n$ such that

$|a_n - \lambda|< \epsilon \hspace{5mm} \forall n > \overline n $

This implies that if $m, n > \overline n$, we also have:

$|a_m - a_n| = |(a_m - \lambda) + (\lambda - a_n)| \leq |a_m - \lambda| + |\lambda - a_| < \epsilon + \epsilon = 2 \epsilon$ hence the result.

The condition is also sufficient. If $(a_n)$ is a Cauchy sequence, we can define an induction sequence of integers $(n_k)$ as follows: $n_1 = 1$, and $n_{k+1}$ is the smallest integer $> n_k$ such that if $m, n > n_{k+1}$, then $|a_m - a_n| < 2^{-k-2}$. Let $I_k$ be the closed interval defined as follows: $[a_{n_k} - 2^{-k}, a_{n_k} + 2^{-k}]$: we have $I_{k+1} \subset I_k$ since $|a_{n_k} - a_{n_{k+1}}| < 2^{-k-1}$, and on the other hand, for $n > n_k$, we have $a_n \in I_k$ by definition. But now it is clear that: $\bigcap\limits_{k=1}^{\infty}I_k = {\lambda}$, where

$\lambda = \underset{k}{sup} (a_{n_k} - 2^{-k}) = \underset{k}{inf} (a_{n_k} + 2^{-k})$

that is to say $\lambda \in I_k \hspace{4mm} \forall k$ and therefore in particular:

$|a_n - \lambda| < 2^{-k-1} \hspace{5mm} \forall n > \overline n_k$.

Giova62 Gds
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