1. ## Convergence Sequence Question

Q: Suppose that {Xn} converges to X. For each n within the set of Natural Numbers, define Yn = (X1+X2+...+Xn)/n. Prove that {Yn} also converges to x.

My solution: So far, I can only think of this as a continuous function problem. I set {Yn} as a function of Y(x). So if Y(x) is continuous, then Y(Xn) -> Y(x).

Am I having the right idea?

Thank you.

KK

Q: Suppose that {Xn} converges to X. For each n within the set of Natural Numbers, define Yn = (X1+X2+...+Xn)/n. Prove that {Yn} also converges to x.
It is sooo late now, I cannot post a complete proof but this is going to be a mess. Hope it will at least help you somehow. (I probably end up deleting this embarassement tomorrow).

By definition of sequences converging to X,
$n\geq N\to |x_n-X|<\epsilon/n$

Prove that $\{y_n\}$ converges to X.
That is,
$n\geq N \to \left| \frac{x_1+x_2+...+x_n}{n}-X \right|<\epsilon$

But,
$\left| \frac{x_1+x_2+...+x_n}{n}-X\right|<\epsilon$
Is,
$\left| \frac{x_1+x_2+...+x_n-nX}{n} \right|<\epsilon$
Thus,
$\left| \frac{(x_1-X)+(x_2-X)+...+(x_n-X)}{n} \right|<\epsilon$
Triangular inequality, and first inequality
$|x_1-X|+|x_2-X|+...+|x_n-X|<\epsilon/n+...+\epsilon/n$
Thus,
$|x_1-X|+...+|x_n-X|<\epsilon$
Which is true.

3. Thanks for the reply. I understand how you sub into the |{Yn} - x| thing, but I don't understand how $|x_1-x|+|x_2 -x|+...+|x_n-x|<\epsilon/n+...$, what I question is how do you know $x_1$ is < $x_n$?

Would you please explain that part? Then I should be good for the rest.

Thank you!!!

KK