1. ## Sequences

Let (xn) and (yn) be two sequences. Let (zn) be the shuffled sequence:
z1=x1, z2=y1, z3=x2, z4=y2.....z(2n-1)=xn, z(2n)=yn...

Prove that (zn) converges, which implies that both (xn) and (yn) converge.

Can someone help?
Thanks.

2. Originally Posted by JaysFan31
Let (xn) and (yn) be two sequences. Let (zn) be the shuffled sequence:
z1=x1, z2=y1, z3=x2, z4=y2.....z(2n-1)=xn, z(2n)=yn...

Prove that (zn) converges, which implies that both (xn) and (yn) converge.

Can someone help?
Thanks.
Counterexample:
Let
x_n = n
y_n = n^2

None of the sequences x, y, z converge.

Do you mean the question to say that "if {z_n} converges then {x_n}, {y_n} converge"?

-Dan

3. Originally Posted by topsquark
Do you mean the question to say that "if {z_n} converges then {x_n}, {y_n} converge"?
Yes.

------
I did not try to solve this problem but I am thinking:
If the sequence {z_n} converges then the odd and even subsequences converge that is,
z_1,z_3,z_5,... -----> Converges
z_2,z_4,z_6,... -----> Converges

But the odd-even subsequences are the sequences x_n and y_n. Thus, {x_n} and {y_n} both converges.

Is that not true?

4. Originally Posted by topsquark

Do you mean the question to say that "if {z_n} converges then {x_n}, {y_n} converge"?
If so, use the fact that a sequence converges iff every subsequence
converges then the result is trivial. So maybe we should conclude that
you don't know this principal?

RonL