# Homework Help: Real Analysis: Stolz–Cesàro Proof

1. Apr 28, 2010

### Gib Z

1. The problem statement, all variables and given/known data
1. Let xn and yn be sequences in R with yn+1 > yn > 0 for all natural numbers n and that yn→∞.
(a) Let m be a natural number. Show that for n > m
$$\frac{x_n}{y_n} = \frac{x_m}{y_n} + \frac{1}{y_n} \sum_{k=m+1}^{n} (x_k - x_{k-1})$$

(b) Deduce from (a) or otherwise that

$$|\frac{x_n}{y_n}| \leq |\frac{x_m}{y_n}| + \sup_{k>m} | \frac{ x_k - x_{k-1} }{ y_k - y_{k-1} } |$$

(c) Assuming $$\frac{x_n-x_{n-1}}{y_n-y_{n-1}} \to 0$$, show $x_n/y_n \to 0$.

(d) Assuming $$\frac{x_n-x_{n-1}}{y_n-y_{n-1}} \to L$$, show $x_n/y_n \to L$.

2. Relevant equations
N/A

3. The attempt at a solution
(a) was fine.

(b) Would the question be more correct to use sup k>m+1 instead, since a k-1 index is in the inside expression?

I'm not sure if this was the right thing to do as the question probably intended the sum to remain unsimplified, but I replaced it with xn-xm.

Using the triangle inequality and that y is strictly increasing so that yn-ym < yn, we get

$$| \frac{x_n}{y_n} | \leq | \frac{x_m}{y_n} | + | \frac{x_n-x_m}{y_n-y_m} |$$

Then I'm not really sure what I can validly do after that.

(c) If (b) is assumed, then if we take the limit of both sides, it reduces to the statement that $$| \frac{x_n}{y_n} | - | \frac{x_m}{y_n} | \leq 0$$ holds true for large n.

I don't think that's the right direction to go, certainly since it seems to imply x is decreasing when that was never given. Don't know what to do.

(d) No idea, but if I had (c) this one would probably be similar.

2. Apr 29, 2010

### Gib Z

Has anyone got any hints? I'm still stuck.

3. Apr 29, 2010

### boboYO

you probably already know this, but the 2nd equation doesn't imply the first because sup(all that stuff) could be smaller than $$\left| \frac{x_n-x_m}{y_n-y_m} \right|$$

The most obvious path is to prove that
$$\sup_{k>m}\left|y_n\frac{\Delta x_k}{\Delta y_k} \right| \geq \left| \sum_{k=m+1}^{n} \Delta x_k\right|$$

and then apply triangle inequality.

First, restate it as an existence question, remove the sup and ask: does there exist a $$k$$ that makes the inequality true?

[STRIKE]An idea:

It reduces to finding $$m+1\leq k \leq n$$ such that

$$\frac{y_n}{\Delta y_k}\geq n-m$$ (1)

and

$$\Delta x_k \geq \frac{|x_n-x_m|}{n-m}$$ (2)

It is easy to to find a $$k$$ that satisfies each condition individually, but I can't see how to find one that satisfies both simultaneously =(
[/STRIKE]
edit: the above approach won't work, I found a counterexample to it.
For c),

to find N such that $$\frac{x_n}{y_n}<\varepsilon$$ for n>N,

first find an m such that $$\frac{ \Delta x_k }{\Delta y_k } < \frac \varepsilon 2$$ for k>m, then use part b) and finish it off.

Last edited: Apr 29, 2010
4. Apr 30, 2010

### Gib Z

My mind started thinking clearly and I figured these out finally. If anyone's interested in the solutions send me a PM and I'll post them here.

boboYO, I didn't see the edit until just now, my final solution for (c) didn't make any use of (b) and was quite long =[ Could you tell me how we could have used (b) to finish it off. When I see it I bet I'll kick myself.

Last edited: Apr 30, 2010
5. May 3, 2010

### lanedance

see what you think of this
let $\Delta x_j = x_j-x_{j-1}$

then, let k>m be the the supremum index, then
$$| \frac{\Delta x_k }{ \Delta y_k }| \geq | \frac{\Delta x_j }{ \Delta y_j }|, \forall j>m$$

rearranging, knowing that $\Delta y_j > 0$, then summing from m+1 upto n:
$$| \frac{\Delta x_k }{ \Delta y_k }| \sum \Delta y_j \geq \sum |\Delta x_j |\geq |\sum \Delta x_j|$$
so
$$\sup_{k>m}| \frac{\Delta x_k }{ \Delta y_k }| \geq |\frac{x_n-x_m}{y_n-y_m}|\geq |\frac{x_n-x_m}{y_n}| \geq |\frac{x_n}{y_n}| - |\frac{x_m}{y_n}|$$

6. May 3, 2010

### lanedance

then for c) choose M such that
$$\frac{x_m-x_{m-1}}{y_m-y_{m-1}} < \epsilon, \forall m>M$$

then for n,m>M
$$|\frac{x_n}{y_n}| \leq |\frac{x_m}{y_n}| + \epsilon$$

$x_m$ is just a number, and as $y_n$ is unbounded as n increases, we should be able to choose N>M so that for all n>N
$$|\frac{x_m}{y_n}|\leq \epsilon$$

then for n>N>M
$$|\frac{x_n}{y_n}| \leq 2\epsilon$$

which should be pretty close....

this part threw me for a bit above
$$\sup_{k>m}| \frac{\Delta x_k }{ \Delta y_k }| \geq |\frac{x_n-x_m}{y_n-y_m}|$$
but i'm thinking maybe the continuous analogue is something like the mean value theorem, except that the maximum gradient must be >= the average

not 100% there is no holes...

Last edited: May 3, 2010