# Prove that xn<= L for all n in N

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• Mar 6th 2011, 07:22 PM
alice8675309
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: ($\displaystyle \forall$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?
• Mar 6th 2011, 08:51 PM
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.
• Mar 7th 2011, 09:01 AM
alice8675309
Quote:

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?
• Mar 7th 2011, 09:32 AM
Plato
If we know that $\displaystyle (x_n)\to L$ then for any $\displaystyle \delta>0$ almost all the terms are in $\displaystyle (L-\delta,L+\delta)$.
Suppose that $\displaystyle x_N>L$ for some $\displaystyle N$.
Then let $\displaystyle \delta=\frac{x_N-L}{2}$.
Now apply the almost all property and get a contradiction.