Proof of Stolz-Cesaro

[Moubinool]

Stolz Cesaro:

Let (x_n) and (y_n) be two sequences with the properties:
a)y_n>0 for any n natural;
b) y_n<y_(n+1) for any n
c) (y_n) is not bounded superior;
d) there exists a=lim(n-->oo) (x_(n+1)-x_n)/(y_(n+1)-y_n)
then there exists lim(n-->oo) x_n/y_n and even more
lim(n-->oo) x_n/y_n=a
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Useful theorem in many sequences problems

[liyi]

No proof?
There's proof in nearly every math analysis books in China.

First consider (x_n-x_{n-1})/(y_n-y_{n-1}) \lim L, L is definite.
for any given \varepsilon > 0, there exists N such that for all n>N it holds

|(x_n-x_{n-1})/(y_n-y_{n-1})-L| < \varepsilon/2

or

L-\varepsilon/2 < (x_n-x_{n-1})/(y_n-y_{n-1})-L < L+\varepsilon/2

it means that for all n>N, all the fractions

(x_{N+1}-x_N)/(y_{N+1}-y_N),
(x_{N+2}-x_{N+1})/(y_{N+2}-y_{N+1}),
...
(x_{n-1}-x_{n-2})/(y_{n-1}-y_{n-2}),
(x_n-x_{n-1})/(y_n-y_{n-1})

are in the range. Since y_n is increasing, and the fractions above have positive denominators, so the range includes

(x_n-x_N)/(y_n-y_N)

because its denominator is the sum of all the denominators of the fractions above and so is the numerator.

hence for n > N it holds that

|(x_n-x_N)/(y_n-y_N) - L| < L+\varepsilon/2

from

x_n / y_n - L = (x_n - L y_n)/y_n + (1-y_N/y_n)((x_n-x_N)/(y_n-y_N) - L)

we conclude that

|x_n / y_n - L| \leq |(x_n - L y_n)/y_n| + |(x_n-x_N)/(y_n-y_N) - L|
= A + B

As we know, B is less than \varepsilon/2 when n>N. Since y_n \lim +\infty, so when n>N'>N, A<\varepsilon/2 too.

this gives |x_n / y_n - L| < \varepsilon for all n>N'.

Now suppose (x_n-x_{n-1})/(y_n-y_{n-1}) \lim +\infty.
Just as the argument above,
we can get x_n-x_{n-1} > y_n - y_{n-1} for n large enough.
y_n\lim +\infty tells us that x_n\lim +\infty and x_n is increasing.

thus apply the theorem we have proved to y_n/x_n

lim y_n/x_n = lim (y_n-y_{n-1})/(x_n-x_{n-1}) = 0

finally we get lim x_n/y_n = +\infty

A similar argument works for (x_n-x_{n-1})/(y_n-y_{n-1}) \lim -\infty.