Give $N\in \mathbb{N}$ for the sequence $a_n=3+(-1)^n$ so that for every $n>N$ and $\epsilon=0.5$...

Solution 1:

To answer the question in the comments, let's assume, for the sake of contradiction, that some $L$ and $N$ exist such that $$n > N \implies |a_n - L| < \frac{1}{2}$$ Fix some odd $m > N$. Then

• $a_m = 2$,
• $m + 1 > N$,
• $a_{m + 1} = 4$.

Using $n = m$, we get $|2 - L| < \frac{1}{2}$, and using $n = m + 1$, we get $|4 - L| < \frac{1}{2}$. Using triangle inequality, $$2 = |(4 - L) - (2 - L)| \le |4- L| + |2 - L| < \frac{1}{2} + \frac{1}{2} = 1,$$ hence $2 < 1$, a contradiction. Thus, no such pair $L$ and $N$ can exist.

If the limit of $a_n$ did exist, then we would have such a pair, but since no such pair exists, the limit of $a_n$ does not exist.