If {an} is bounded sequence with exactly one subsequential limit a, the {an} converges to a.
Could you help me understand what bounded sequence with one subsequential limit is? I don't understand the question. Thanks very much in advance
If {an} is bounded sequence with exactly one subsequential limit a, the {an} converges to a.
Could you help me understand what bounded sequence with one subsequential limit is? I don't understand the question. Thanks very much in advance
so by definition lim {a_n_k} = a so for every e > 0 there exists n* in N such that for all n_k>= n* we have with any arbitrary a_n_k being subsequence of a_n:
|a_n_1 - a| < e/n for all n >= n1*
|a_n_2 - a| < e/n for all n >= n2*
|a_n_3 - a| < e/n for all n >= n3*
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|a_n_k - a| < e/n for all n_k >= nk*
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we need to prove that |a_n - a| < e
we know what a_n is bounded so by definition there exist some real number M such that |a_n| < M
since a_n_k is a subsequence of a_n so we have: n_1 < n_2 < n_3,...with i <= n_i , i in N.
Pick n* = min (n_1, n_2, n_3, ....) so for all n >= n* we have
|an_1 + an_2 + an_3 +...+ an_k+.....-n.a| <e
I started getting confused from this. Please help *_*
By assumption we know that every convergent subsequence of (a_n) converges against the same number a.
Furthermore |a_n|<C for all n.
We need to show that a_n also converges.
Assume that this is not the case. In particular a_n does not converge against a. Then there is an e>0 , so that there is a subsequence a_n_j of a_n with |a_n_j-a|>=e for all j.
But a_n_j is also a bounded sequence and any bounded sequence has a convergent subsequence. So a_n_j has a subsequence which converges, but it can't converge against a, because its distance from a is always greater or equal than e.
This subsequence is also a convergent subsequence of a_n. => Contradiction to our assumption.