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

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- Oct 1st 2010, 05:08 AMtommy11Subsequential limit problem
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 - Oct 1st 2010, 06:45 AMIondor
A sequence is "bounded", if there is a constant C>0 , such that

|a_n|< C for all n.

A "subsequential limit" of a sequence is a limit of a subsequence a_n_k of a_n. - Oct 1st 2010, 10:36 AMtommy11
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*

.

.

|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 *_* - Oct 1st 2010, 10:48 AMPlato
- Oct 1st 2010, 12:34 PMIondor
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. - Oct 1st 2010, 12:38 PMIondor
- Oct 1st 2010, 03:37 PMPlato