# Subsequence of the sequence $\frac{1}{n}$

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 $$|a_m - a_n| < 2^{-k-2}$$.

• What have you found for $n_k$? Commented Jul 17, 2023 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. Commented Jul 17, 2023 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 $|a_m - a_n| < 2^{-3}$ Commented Jul 17, 2023 at 19:48

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 $$n_{k+1}=2^{k+2}-1.$$

• In my opinion, $n_1 \neq 1$ Commented Jul 17, 2023 at 20:10
• The assumption is $n_1=1.$ The condition is not true for $k=0$ under that assumption, but $n_1=1$ is given, so I assumed the condition on $n_{k+1}$ was for $k\geq 1.$ @Giova62Gds Commented Jul 17, 2023 at 20:17
• In any event, there is no "opinion" about it. It is a given that $n_1=1,$ so if you wanted differently, you'd need to change your question. @Giova62Gds Commented Jul 17, 2023 at 20:20
• sorry, for my bad english. Commented Jul 17, 2023 at 22:05
• Why exactly $< 1/(n_{k+1} + 1)$ ? Commented Jul 18, 2023 at 15:54

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:

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