# Proving a limit using precise definition

Make a conjecture about $\lim_{n \to \infty} s_k$ , and prove your conjecture.

$s_k= \begin{cases} k, & \text{if$k$is even} \\[2ex] \frac{1}{k}, & \text{if$k$is odd} \end{cases}$

I have come to the conclusion that the limit does not exist? But I am unsure how to prove this using the precise definition of the limit. Do I choose a concrete number for epsilon and show that the limit will exceed that?

## 4 Answers

Suppose it did converge (which I take to mean to a finite limit), call it $L$. That would mean for a fixed epsilon, there would be some $N$ s.t for any $k>N$, we have $$L-\epsilon<k<L+\epsilon$$ but this is certainly false for either $\lceil L+\epsilon \rceil$ or $\lceil L+\epsilon \rceil+1$ depending on the parity of $\lceil L+\epsilon \rceil$

Assume the limit exists Assume the limit is $p$

By necessity, since we assumed the limit exists, $\forall \epsilon>0, \exists N s.t. n\geq N \rightarrow |s_n - p| < \epsilon$

right?

So we can set an $\epsilon$, say $\epsilon =0.05$ and since we assumed the limit exists, there is some $N_{0.05}$ such that the property above holds.

However, we can show that it won't hold for all of $s_{2k}$ (because they strictly increase by 2 every time)

Thus, if the property holds for $n=2l$, $|2l-p|<0.05$ then we know that for $n' = 2l+2$ $|(2l+2)-p| \geq \min\{|2|\pm |2l-p|\} \geq 1.95$ must be at least $1.95$ and thus cannot be $\leq 0.05$

So we get a contradiction and thus the limit must not exist (because the property doesn't hold $\forall n\geq N_{0.05}$ which is just one possible $\epsilon$

HINT

Assume for the sake of contradiction that $$\lim_{k \to \infty} s_k = s.$$ Then, $\forall \epsilon > 0 \exists M > 0$ such that $\forall k > M$ we have $|s_k - s| < \epsilon$.

What happens for even $k>L$?

If $L=\lim_{n\to \infty}s_n$ exists then $\{s_n: n\in \mathbb N\}$ is a bounded set. That is, for some $M>0$ we have $\{s_n:n\in \mathbb N\}\subset [-M,M].$

Because we can take $n_0\in \mathbb N$ such that $\forall n>n_0\;(|x_n-L|<1).$ So let $M=1+|L|+\max \{|x_j|:j\leq n_0\}.$

But in the Q, $\{s_n:n\in \mathbb N\}$ is not a bounded set. Because for any $M>0$ there exists even $k>M,$ so $s_k=k\not \in [-M,M]$, so $\{s_n:n\in \mathbb N\}\not \subset [-M,M].$