# First time poster - Hard limit proof!

1. Nov 14, 2012

### kenb1993

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

Suppose an and bn are sequences where bn is increasing and approaching positive infinity. Assume that lim n->∞ [ ( an+1 - an ) / ( bn+1 - bn) ]= L, where L is a real number. Prove that lim n->∞ [ an / bn ] = L.

2. Relevant equations

Limit theorems

3. The attempt at a solution

I tried making an argument with the basis that both sequences have the same number of terms, so L is the limit of (an+1)/(bn+1) and thus also (an/bn)

Last edited: Nov 14, 2012
2. Nov 14, 2012

### jbunniii

Given $\epsilon > 0$, we can find $N \in \mathbb{N}$ such that
$$(L - \epsilon)(b_{n+1} - b_n) \leq a_{n+1}-a_n \leq (L + \epsilon)(b_{n+1} - b_n)$$
for all $n \geq N$. This is true by definition of the limit and by the fact that $b_{n+1} > b_n$.

Big hint: now try summing this inequality starting at $n = N$ up to some arbitrary value, say $n = M$. Can you conclude anything as $M \rightarrow \infty$?

3. Nov 15, 2012

### kenb1993

So starting at a certain value N in the sequence, when you take the infinite sum of the inequality an and an+1 would tend to positive infinity along with bn+1 but I can't tie an argument together to actually prove the result.

4. Nov 15, 2012

### HallsofIvy

That doesn't make any sense. Both sequences are infinite. That don't have a "number of terms". It is however, true that the limits of the two sequences you mention are the same but that really doesn't matter. The sequence (an+1- an)/(bn+1- bn[/b]) is NOT (an+1/bn+1)- (an/bn).

5. Nov 15, 2012

### jbunniii

Try writing out explicitly what you get if you sum the inequality in my previous post from $n = N$ to $n = M-1$. Notice that all three terms will "telescope." What is the result? Write it out in detail.

6. Nov 15, 2012

### kenb1993

After manipulation were left with (L−ϵ)(1-bN/bM) + aN/bM≤aM/bM≤(L+ϵ)(1-bN/bM) + aN/bM.

Then would bN/bM converge to zero and we are done?

Last edited: Nov 15, 2012
7. Nov 15, 2012

### jbunniii

Well, you need both $b_N/b_M$ and $a_N/b_M$ to converge to zero as $M \rightarrow \infty$. Do they?

8. Nov 15, 2012

### kenb1993

I think that's safe to assume without proof since bn is increasing. Thank for your help.

9. Nov 15, 2012

### jbunniii

It's not because $b_n$ is increasing. It's because $b_n$ is diverging to positive infinity (a fact you haven't used yet). Therefore $1/b_n$ is converging to...?

10. Nov 15, 2012

### kenb1993

Zero haha. Thanks. xn