help with question about sequences??

Question

Show that if xn and yn are convergent sequences of real numbers, then the sequence zn= max(xn,yn) is also convergent...

Answer 1

We consider two cases. First, suppose that the two sequences converge to the same limit, L. Then ∀ε>0, ∃(δ_x, δ_y)∈N×N : ∀n>δ_x, |x_n-L| < ε and ∀n>δ_y, |y_n-L| < ε. Let δ = max(δ_x, δ_y). Since for all n>δ, n>δ_x and n>δ_y, it follows that for all n>δ, |x_n-L| < ε and |y_n-L| < ε. But since ∀n, either z_n=x_n or z_n=y_n, this means that ∀n>δ, |z_n-L|<ε. And since we can find this δ for all ε>0, it follows that z_n converges to the limit L

Now, suppose that the two sequences converge to different limits. Without loss of generality suppose [n→∞]lim x_n > [n→∞]lim y_n. Then there exists δ such that ∀n>δ, x_n>y_n, and thus ∀n>δ, z_n=x_n. That is, all but a finite prefix of z_n is identical to x_n. Since changing a finite number of terms at the beginning of a series does not affect either convergence or limits, it follows that z_n converges, to the same limit as x_n (if it were the case instead [n→∞]lim y_n > [n→∞]lim x_n, then it would converge to the same limit as y_n).

Answer 2

arghh, you need a proof? i proved zero once but didn't get much further. They're both convergent. Proving them graphically is my favorite way to prove this though most teachers don't take a liking to it on the final pencil/paper exam. Though i had one who did. Life was good.

Plug in dummy values to prove. cover all those then generalize them. Voila. your answer and proof.