Let t and s be real numbers with $t. Suppose that as n→∞, a2n→t and a2n−1→s.
True or False: It must be that lim infan=t and lim supan=s and (an) has no other limit points.
Prove the statement or give a counterexample.
I think true, based on the following example, which I adapted from a wikipedia image (where s=1, t=-1 and a2n−1 is the pink sequence and a2n is the light blue sequence):
Hence, all of what the statement said is true... but i don't know how to prove it.
Answer
I will use the following definition of lim sup and lim inf:
Let E be the set of subsequential limits of (an). Then lim supn→∞an:=supE and lim infn→∞an:=infE.
It suffices to show that in the setting of your question, E={t,s}.
Clearly t,s∈E. Now suppose (ank) is a subsequence of (an). The sequence of indices nk must either contain infinitely many even numbers or infinitely many odd numbers (or both).
Edit: This implies that (ank) contains a further subsequence that either converges to s or to t. Thus if (ank) converges, it must be to either s or t. (But it might not converge.)
So E={t,s}.
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