Let be a sequence in
Ak ≡ sup so that
limsupn→∞an = limn→∞An, the An being a decreasing sequence.
- Show that in all cases, there exists Bn < An such that Bn is increasing
and limn→∞Bn = λ.
- Explain why, in all cases there are infinitely many k such that ak ∈
Hint: If for all k ≥ m > n, ak ≤ Bn, then ak < Bm also and
≤ Bm < Am contrary to the definition of Am.
- Explain why there exists a subsequence such that lim
k→∞ank = λ.
- Show that if γ ∈ and there is a subsequence
k→∞ank = γ, then γ ≤ λ.
This shows that limsupn→∞an is the largest in such that some
subsequence converges to it. Would it all work if you only assumed that
−∞ for infinitely many n? What if an = −∞ for all n large enough? Isn’t this case
fairly easy? The next few problems are similar.