Question: Suppose that the sequence $(a_n)$ converges to 7. Prove that there exists a subsequence $(a_{n_k})$ such that $|a_{n_k} − 7| < \frac{1}{10^k}$ for all k. (In other words, the subsequence converges to
7 rapidly.)
I currently have:
Let $\epsilon$ be an arbitrary positive real number. Assume that the sequence $(a_n)$ converges to 7, $|a_n − 7| < \epsilon$. By Theorem, the subsequence $(a_{n_k})$ of $(a_n)$ converges to 7 as well, $|a_{n_k} − 7| < \epsilon$.
The theorem used is just that the subsequence will converge to the same point as the original sequence.
From here, I don't know where to go. Should I have chosen $\epsilon = \frac{1}{10^k}$?
Best Answer
For $ε=\frac{1}{10}$ there exists $l_1\in\mathbb{N}$ such that for every $n\geq l_1$ we have $|a_n-7|<\frac{1}{10}$. We choose and arbitrary $n_1\geq l_1$ and we have $|a_{n_1}-7|<\frac{1}{10}$. For $ε=\frac{1}{10^2}$ there exists $l_2\in\mathbb{N}$ such that for every $n\geq l_2$ we have $|a_n-7|<\frac{1}{10^2}$. We choose $n_2\geq l_2$ with $n_2>n_1$ and we have $|a_{n_2}-7|<\frac{1}{10^2}$. Such $n_2$ exists because the relation $|a_n-7|<\frac{1}{10^2}$ holds for every $n\geq l_2$ and so we can find a natural number greater than $n_1$. Following, by induction you can show that there exists a sequence $n_1<n_2<\cdots<n_k<\cdots$ such that $|a_{n_k}-7|<\frac{1}{10^k}$ for every $k\in\mathbb{N}$ and so we have a subsequence that does the job.