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Given a sequence $\{a_n\}_{n=1}^{\infty}$ and $c>0$ such that $a_{n+1} - a_n>c$ for every $n$, prove $\lim_{n\to\infty} a_n=\infty$.

I proved $a_n$ is monotonic increasing, but I'm having hard time proving it's unbounded.

Any ideas?

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From the fact that $a_{n+1} > c+a_n$, you should prove (by induction) that $a_{n+1} > n\cdot c + a_1$.

Once you have that, proving that the sequence is unbounded should be a piece of pie, since for every $M\in\mathbb R$, you can set $n=\frac{M-a_1}{c}$ and get that $a_{n+1} > M.$

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  • $\begingroup$ Yes, I've seen a similar solution, thought there was another way. Could you please explain how did you think of the inequality need to be proven ? $\endgroup$ – Itay4 Feb 14 '17 at 13:45
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    $\begingroup$ @Itay4 Well, after you just rearange $a_{n+1} - a_n>c$ into $a_{n+1} > c + a_n$, you can see that this equality actually means that "after each step, you move at least $c$ to the right". That's only one logical leap from thinking OK, so after $2$ steps, I am $2c$ to the right, and after $1000$ steps, I am $1000c$ to the right, so after $n$ steps, I will be $nc$ to the right". $\endgroup$ – 5xum Feb 14 '17 at 13:46
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    $\begingroup$ @Itay4 Then, you just put that into equation form and rigorously prove it. $\endgroup$ – 5xum Feb 14 '17 at 13:47
  • $\begingroup$ Got it, thanks ! $\endgroup$ – Itay4 Feb 14 '17 at 13:49
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Prove $a_{n+1}>a_1+nc$ using induction or telescoping.

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you can see that

$$ a_n-a_0= \sum_{i=1}^{n } a_i-a_{i-1}> c\sum_{i=1}^{n} 1 =cn $$

ie $$\infty = \lim nc+a_0 < \lim a_n $$ hence $a_n\to \infty$

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  • $\begingroup$ Nice ! I know that weak inequality is preserved in limits, is it the same on strong one? $\endgroup$ – Itay4 Feb 14 '17 at 14:04
  • $\begingroup$ What do you mean by weak inequality in this context? Please clarify $\endgroup$ – Guy Fsone Feb 16 '17 at 11:41
  • $\begingroup$ I know that if $a\ \geq b$ then $lima\ \geq lim b$. Is it the same with "$>$" ? $\endgroup$ – Itay4 Feb 16 '17 at 14:49
  • $\begingroup$ $a<b\implies \lim a\le\lim b $. fpr instance $x>1\implies x< x^2$ but the limits coincide as $x\to 1.$ $\endgroup$ – Guy Fsone Feb 17 '17 at 11:34
  • $\begingroup$ Got it, thanks! $\endgroup$ – Itay4 Feb 17 '17 at 13:42

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