# If a sequence is eventually bounded then it is bounded

1. Jun 16, 2015

### pixelate

1. The problem statement, all variables and given/known data

Hi, I've been solving Calculus Deconstructed by Nitecki and I've been confused by a particular lemma in the book. Namely:

If a sequence is eventually bounded, then it is bounded:
that is, to show that a sequence is bounded, we need only find a number
γ ∈ R such that the inequality

$$|x_i| < \gamma \ \text{holds for all i ≥ K, for some K}$$

2. Relevant equations
3. The attempt at a solution

However if we consider the sequence $\tan {\frac{\pi}{n}}$, with n starting at 1. For terms when n is very high, we can find some value for $\gamma$ for which the condition holds and thus the sequence is bounded (according to the lemma). Yet at n=1, $\tan {\frac{\pi}{n}} = \infty$, so the sequence cannot be bounded.

My questions are:

1. Is my reasoning at all correct in the first place ?

2. It is easy to see that my sequence does have a lower bound but no upper bound, so does the lemma only refer to one bound (But doesn't bounded mean bounded in both directions?)

2. Jun 16, 2015

### Ray Vickson

You don't have a valid sequence, so that is not a counterexample to anything.

Anyway, at $n = 1$ we have $\tan(\pi) = 0$, so that is OK. The trouble occurs at $n = 2$.

But, I say it again: you don't have a valid sequence of real numbers with this example.

3. Jun 16, 2015

### pixelate

.Oh I see, thats very illuminating thank you. Of course at n=2, the the element of the sequence is undefined and it is not infinite at all. I feel silly for making such a basic mistake now.