If a (x_n) --> M, then (y_m=(x_1+...+x_m)/m) --> M

Skerven · #1 $Old$ November 3rd, 2009, 06:33 AM

$(x_{n})$ and $y_{m} = \frac{x_{1}+...+x_{m}}{m}$

I need to show that $\lim_{n\to\infty}(x_{n}) = M \rightarrow \lim_{m\to\infty}(y_{m}) = M$ .

I started by writing out the definition of convergence for $(x_{n})$
$\forall \epsilon > 0, \exists k(\epsilon) : \forall n > k(\epsilon), |x_{n} - M| < \epsilon$

Then I discovered that
$x_{n} = y_{n} + (n - 1) * (y_{n+1} - y_{n})$
But I don't know how to show the convergence of $(y_{n})$

Plato · #2 $Old$ November 3rd, 2009, 09:01 AM

Quote:

Originally Posted by Skerven $View Post$

$(x_{n})$ and $y_{m} = \frac{x_{1}+...+x_{m}}{m}$
Show that $\lim_{n\to\infty}(x_{n}) = M \rightarrow \lim_{m\to\infty}(y_{m}) = M$ .

I find it easier to do this by using two cases.
Case I. Suppose that $M=0$ . Then use the notation $S_N = \sum\limits_{k = 1}^N {x_k }$ .
If $\varepsilon > 0$ find a $K$ such $n\ge K$ implies that $\frac{{\left| {S_K } \right|}}{n} < \frac{\varepsilon }{2}\;\& \,\left| {x_n } \right| < \frac{\varepsilon }{2}$ .
Then $\left| {y_n } \right| = \left| {\frac{{S_n }} {n}} \right| = \left| {\frac{{S_n - S_K + S_K }} {n}} \right| \leqslant \frac{1} {n}\sum\limits_{k = K + 1}^n {\left| {x_k } \right|} + \frac{{\left| {S_K } \right|}} {n} < \frac{1} {n}\left( {n - K + 1} \right)\frac{\varepsilon } {2} + \frac{\varepsilon } {2}$ .

Case II. $M\ne 0$ In that case let $z_n=x_n-M$ and apply case I.

Skerven · #3 $Old$ November 3rd, 2009, 10:21 AM

Quote:

Originally Posted by Plato $View Post$

If $\varepsilon > 0$ find a $K$ such $n\ge K$ implies that $\frac{{\left| {S_K } \right|}}{n} < \frac{\varepsilon }{2}$

How do you know that you can find such a $K$ ? It seems like you are assuming what you are trying to prove.

Edit: Ok, so you are choosing a K that satisfies the second inequality, justified by assumption.

Then you are increasing K until you find a range of n that satisfied the first inequality. Got it.