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: ($\displaystyle \forall$n$\displaystyle \in$N) $\displaystyle \exists$k such that x_m>x_n and (xn)->L which means: $\displaystyle \forall$$\displaystyle \epsilon$>0 $\displaystyle \exists$k such that for all n$\displaystyle \geq$k we have |x_n-L|<$\displaystyle \epsilon$

and we need: x_n≤L

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