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.
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?