# Subsequential limit problem

• Oct 1st 2010, 04:08 AM
tommy11
Subsequential 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, 05:45 AM
Iondor
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, 09:36 AM
tommy11
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*
.
.
.
.
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

• Oct 1st 2010, 09:48 AM
Plato
Quote:

Originally Posted by tommy11
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

Because every sequence by definition is a subsquence of itself, there really nothing to prove.
• Oct 1st 2010, 11:34 AM
Iondor
Quote:

Originally Posted by tommy11
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*
.
.
.
.
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

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, 11:38 AM
Iondor
Quote:

Originally Posted by Plato
Because every sequence by definition is a subsquence of itself, there really nothing to prove.

But not every subsequence needs to converge.
The assumption is only that every convergent subsequence of (a_n) has the same limit.
• Oct 1st 2010, 02:37 PM
Plato
Quote:

Originally Posted by Iondor
But not every subsequence needs to converge.
The assumption is only that every convergent subsequence of (a_n) has the same limit.

And you proof shows that every subsequence does in fact converge.