First time poster - Hard limit proof

[kenb1993]

Homework Statement

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

Homework Equations

Limit theorems

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 (a_n+1)/(b_n+1) and thus also (a_n/b_n)

 

[jbunniii]

Homework Statement

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

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 $M \rightarrow \infty$?

 

[kenb1993]

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

 

[HallsofIvy]

Homework Statement

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

Homework Equations

Limit theorems

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 (a_n+1)/(b_n+1) and thus also (a_n/b_n)

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 (a_n+1- a_n)/(b_n+1- b_{n[/b]) is NOT (an+1/bn+1)- (an/bn).}

 

[jbunniii]

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

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.

 

[kenb1993]

After manipulation were left with (L−ϵ)(1-b_N/b_M) + a_N/b_M≤a_M/b_M≤(L+ϵ)(1-b_N/b_M) + a_N/b_M.

Then would b_N/b_M converge to zero and we are done?

 

[jbunniii]

After manipulation were left with (L−ϵ)(1-b_N/b_M) + a_N/b_M≤a_M/b_M≤(L+ϵ)(1-b_N/b_M) + a_N/b_M.

Then would b_N/b_M converge to zero and we are done?

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

 

[kenb1993]

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

 

[jbunniii]

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

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...?

 

[kenb1993]

Zero haha. Thanks. x_n