If a sequence is eventually bounded then it is bounded

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Homework Statement

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}

Homework Equations

3. The Attempt at a Solution
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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?)

Thanks for reading!

 

[Ray Vickson]

Homework Statement

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}

Homework Equations

3. The Attempt at a Solution
[/B]
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?)

Thanks for reading!

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.

 

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

.Oh I see, that's 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.