Prove the existance of a subsequence such that

[hages]

Homework Statement

Notation:
|a_n| is the absolute value of a_n.
(s_n) signifies a sequences; s_n signifies the value of the sequence at a particular n.

Problem:
Let n, k be arbitrary elements in N.
Let (a_n) be a sequence such that lim inf |a_n| = 0. Prove that there is a subsequence (an_k) such that the series, SIGMA an_k from k=1 to +infinity converges.

Homework Equations

None.

The Attempt at a Solution

I am just at a complete loss for how to solve this problem.

Here's my outline:
Let s_n = SIGMA |a_k| from k = 1 to n. Since for all n, |a_n| => 0, (s_n) is nondecreasing.

Now if I can show that (s_n) is bounded, I'd know that (s_n) converges and thus would know there is a convergent subsequence (sn_j) where sn_j = SIGMA |an_k| from k=1 to j. Thus we'd say that SIGMA |an_k| from k=1 to j would be convergent. And thus we'd have that SIGMA an_k from k=1 to +infinity would be convergent because absolutely convergent series are convergent.

But following this route means I have to prove that (s_n) is bounded and that is where I happen to be stuck. Can anyone help me out?

 

[lanedance]

ok so I'm fairly new to this stuff (disclaimer), but we need to find a subsequence s_k (changed notation as the n was confusing) is

as the sequence a_n converges to zero for any e>0, there exists N>0 such that for all n>N, |a_n|<e

so why not pick each k such that the kth term will be less than some convergent sequence we know of, at least for some k>K. this should guarantee the convergence of s_k

so how about |a_nk|<1/k², can pick each of these, due to the convergence of a_n to zero and this should absolute convergence

ie sum |a_k|< A + sum (1/k^2)
where A is some bounded constant

 

[fmam3]

ok so I'm fairly new to this stuff (disclaimer), but we need to find a subsequence s_k (changed notation as the n was confusing) is

as the sequence a_n converges to zero for any e>0, there exists N>0 such that for all n>N, |a_n|<e

so why not pick each k such that the kth term will be less than some convergent sequence we know of, at least for some k>K. this should guarantee the convergence of s_k

so how about |a_nk|<1/k², can pick each of these, due to the convergence of a_n to zero and this should absolute convergence

ie sum |aSUB]k[/SUB]|< A + sum (1/k^2)
where A is some bounded constant

I don't think its entirely correct to say that the sequence $$(a_n)$$ converges to zero. Note that it is only given that $$\liminf |a_n| = 0$$ and we have $$\lim |a_n|= 0$$ if and only if $$\liminf |a_n| = \limsup |a_n|$$.

 

[lanedance]

i read it as "lim n goes to inf -> |a_n|->0"

but in your description, hopefully we only need a small change

lim inf{|a_n|} implies that for any e>0, there's an infinite amount of a_n, such that |a_n| < 0+e, thus implying the existence of a subsequence convergent to zero anyway?

 

[fmam3]

i read it as "lim n goes to inf -> |a_n|"

but in your description, hopefully we only need a small change

Actually, liminf and limsup (i.e. limit inferior and limit superior) are precisely defined as follows. Let $$(s_n)$$ be a sequence of real numbers. Then define $$u_N = \inf\{s_n : n > N\}$$. Then $$\liminf s_n = \lim_{N \to +\infty}u_N$$. Limit superior is defined analogously. There are many ways of defining liminf and limsup but I'm just providing one of the definitions here.

lim inf{|a_n|} implies that for any e>0, there's an infinite amount of a_n, such that |a_n| < 0+e

If you interpret liminf like that, then there's no difference between the liminf and the usual limit.

For instance, if we take a sequence of numbers like $$s_{2n} = +1, s_{2n +1} = -1$$, then clearly the sequence $$(s_n)$$ does not have a limit. But it is clear that $$\liminf = -1, \limsup = +1$$.



Actually, the idea that you'd showed in the earlier post is good but the argument could be made more clear and precise. (see next post)

 

[fmam3]

To clear up the OP's problem statement:

Let $$(a_n)$$ be a sequence of reals such that $$\liminf |a_n| = 0$$. Prove that there is a subsequence $$(a_{n_k})$$ such that the series $$\sum_{k=1}^{\infty} a_{n_k}$$ converges.

Let $$\forall \varepsilon > 0$$. Since $$\liminf |a_n| = 0$$, there $$\exists N > 0$$ such that for $$n \geq N$$ we have $$|a_n| < \varepsilon$$.

In particular, for $$n_1 > N$$, we have $$|a_{n_1}| < 1$$. And again, for $$n_2 > n_1 > N$$, we have $$|a_{n_2}| < \frac{1}{2^2}$$. (I'm omitting the induction argument here because I'm lazy) but eventually, you'll have that $$|a_{n_k}| < 1/k^2$$ for $$\forall k \in \mathbb{N}$$. And since the RHS is a p-series, which converges, and thus by the Comparison Test, we have that $$\sum |a_{n_k}|$$ converges. Since absolute convergence implies convergence, it follows that $$\sum a_{n_k}$$ converges.

 

[fmam3]

Actually, an even easier argument could be made as follows. Since $$\liminf |a_n| = 0$$ and since the liminf is the infimum of the set of subsequential limits, there exists a subsequence $$(|a_{n_k}|)$$ such that $$\lim_{k \to \infty} |a_{n_k}| = 0$$. This means that $$\forall \varepsilon > 0, \exists N > 0$$ such that $$\forall k > N$$, we have $$| |a_{n_k}| | = |a_{n_k}| < \varepsilon$$. Repeat the same argument that I'd provided above.

But note that in the first argument above, I never used any properties of the set of subsequential limits; in fact, I just went by definition of what liminf means and from that, I had to construct my subsequence. But in this argument, I already "know" there will be a subsequence that converges to the limit zero. And from this, we can use standard limit properties to construct what we'd desired. It's quite different in approach, IMHO, even though both use a similar argument and I would think that the subsequential limit approach is much more sophisticated (i.e. you need to actually prove that if $$S$$ is the set of subsequential limits of the sequence $$(s_n)$$, then $$\inf S = \liminf s_n$$.)

 

[hages]

Thank you all. As you can see from my original post, I hadn't thought of actually constructing a subsequence with the desired property. I appreciate the help.