# Proving $\lim_{n\to\infty} a_n=\infty$

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?

## 3 Answers

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 $a_{n+1} > M.$

• 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 ? – Itay4 Feb 14 '17 at 13:45
• @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". – 5xum Feb 14 '17 at 13:46
• @Itay4 Then, you just put that into equation form and rigorously prove it. – 5xum Feb 14 '17 at 13:47
• Got it, thanks ! – Itay4 Feb 14 '17 at 13:49

Prove $a_{n+1}>a_1+nc$ using induction or telescoping.

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 $a_n\to \infty$

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