Thread: Prove that xn<= L for all n in N

Assume that (x_n) is a sequence satisfying: For every n∈N there is an integer m>n such that xm >xn. Also assume that (x_n) converges to L.

Prove that xn≤L for all n∈N.

I started out stating what I had:

Have: ( n N) k such that x_m>x_n and (xn)->L which means: >0 k such that for all n k we have |x_n-L|<

and we need: x_n≤L


How would I put that info into a proof and would it be by contradiction?

Assume there is an n such that xn>L. Then show there are infinitely many m>n such that xm>xn. This contradicts convergence of xn to L.

Originally Posted by Tinyboss

Assume there is an n such that xn>L. Then show there are infinitely many m>n such that xm>xn. This contradicts convergence of xn to L.

How would this be shown if we wanted strictly less than?

If we know that then for any almost all the terms are in .
Suppose that for some .
Then let .
Now apply the almost all property and get a contradiction.