Prove that an increasing sequence with a numerical limit is bounded.

I found the following proof referred to a convergent sequence, Can I use the same for this case (an increasing sequence)? I do not understand very well what is "$ C $" . What happens if the sequence is decreasing? Is there a better way to prove the theorem?


Let $s_n$ the sequence that converges to $l \in \mathbb{R}$. We take $\varepsilon =1 $ in the definition of limit, then exists a $n_o \in \mathbb{N}$ such that $|s_n-l|<\varepsilon \ , \forall \ n \geq n_o$. If $C=\max\{1,|s_1-l|,|s_2-l|\dots|s_{n_0-1}-l|\}$ then $|s_n-l|\leq C,$ it means $l-C\leq s_n\leq l+C $ so $s_n$ is bounded.

  • $\begingroup$ What is a numerical limit? Does it mean a finite limit? $\endgroup$ – B. Mehta Dec 28 '17 at 20:45
  • $\begingroup$ @B.Mehta: I think so. $\endgroup$ – Bernard Dec 28 '17 at 20:46
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    $\begingroup$ This proof is valid in general. $\endgroup$ – Gödel Dec 28 '17 at 20:48
  • $\begingroup$ easier proof: all terms of the sequence are between $s_1$ and $l$, so the sequence is bounded. $\endgroup$ – ThePortakal Dec 28 '17 at 20:55

I don't think there can be a simpler proof.

Let me summarise the proof: by definition of the limit $\ell$, there exists an integer $n_0$ such that $$|s_n-\ell| < 1\quad\forall n\ge n_0.$$ For the values of $n<n_0$ it may or may not be that $|s_n-\ell|\le 1$. But there's a finite number of such $n$s anyway, so we may consider the greatest of all these numbers: $$|s_0-\ell|,\, |s_1-\ell|, \dots,\,|s_{n_0-1}-\ell|,\, 1,$$ which we denote $C$.

Now it is by construction that we have $|s_n-\ell|\le C$ for any $n$, since it is always no greater than one of the numbers in the list.

  • $\begingroup$ Thank you, so if I want to prove that an decreasing sequence with a limit is bounded, can I use the same proof? $\endgroup$ – B. David Dec 28 '17 at 22:21
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    $\begingroup$ Absolutely. Note I nowhere used the monotonicity of the sequence: all convergent sequences are bounded. $\endgroup$ – Bernard Dec 28 '17 at 22:24

With $\epsilon=1, \exists m \in \Bbb N \; $, such that

$$s_0\le s_n\le l+1$$ for $n>m.$

thus $\max (s_0,s_1,...s_m) =s_m$

and $$\forall n\in\Bbb N \;\; s_0\le s_n\le \max(s_m,l+1) .$$



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