Real Analysis: Stolz–Cesàro Proof

[Gib Z]

Homework Statement

1. Let x_n and y_n be sequences in R with y_n+1 > y_n > 0 for all natural numbers n and that y_n→∞.
(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.

Homework Equations

N/A

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 x_n-x_m.

Using the triangle inequality and that y is strictly increasing so that y_n-y_m < y_n, 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.

 

[Gib Z]

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

 

[boboYO]

(b) Deduce from (a) or otherwise that

|\frac{x_n}{y_n}| \leq |\frac{x_m}{y_n}| + \sup_{k&gt;m} | \frac{ x_k - x_{k-1} }{ y_k - y_{k-1} } |



\left| \frac{x_n}{y_n} \right| \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.

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&gt;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}&lt;\varepsilon for n>N,

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

 

[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.

 

[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&gt;m

rearranging, knowing that \Delta y_j &gt; 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&gt;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}|

 

[lanedance]

then for c) choose M such that
\frac{x_m-x_{m-1}}{y_m-y_{m-1}} &lt; \epsilon, \forall m&gt;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&gt;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...