Questions on metric spaces?

Question

1) GIve an example of a sequence {s_n} (n=1 to infinity) in L^2 s.t. sum from n=1 to infinity of abs(s_n) = infinity

2) Let {a_n} be in L^2. Prove that sum (n=1 to infinity) a_n/n is absolutely convergent.

For 1), i'm still trying to formulate something

For 2) I'm pretty sure this is just an application of schwarz's inequality.  We let a_n just be itself and we let 1/n be the other sequence, t_n.  All we have to show is that 1/n is in L^2. And we are done.  Which is to say that the sum from n=1 to infinity of 1/n^2 is less than infinity, which is true, so we are done.

So, therefore the product of those sequences is absolutely convergent.  That right?

Ben · Accepted Answer

1) You should define L^2 for the problem.  I believe, correct me if I'm wrong, that you're referring to the space of sequences for which ∑ (n=1 to inf)  |s_n|^2 is finite.
An example of a sequence in L^2 which is not in L^1 is
s_n = 1/n
The sum is infinite, but the sum of the squares is π^2/6

2)
Well it seems you do have the essence of the proof; indeed the Cauchy-Schwartz inequality results from the fact that L^2 satisfies the conditions of a norm (i.e. satisfies the triangle inequality).   However, it seems you need a way to show that ∑(1 to inf) 1/n^2 is finite.  The most obvious way to prove this fact is to use the "integral test" from calc 2.

In the most formal manner of the test, we have:
1/n^2 = [n - (n - 1)]*1/n^2 = ∫(n-1 ... n) 1/n^2 dt 
because 1/t^2 is strictly monotonically decreasing for positive t,
∫(n-1 ... n) 1/n^2 dt < ∫(n-1 ... n) 1/t^2 dt

By applying this inequality to the following sum, we have:
∑(n=2 to inf) 1/n^2 < ∑(n=2 to inf) ∫(n-1 ... n) 1/t^2 dt = ∫(1 ... inf) 1/t^2 dt
evaluating that improper integral, we get that
∑(n=2 to inf) 1/n^2 < 1
Thus,
∑(n=1 to inf) 1/n^2 = 1/1 + ∑(n=2 to inf) 1/n^2 < 1 + 1 = 2 < inf

Thus, the sum converges.  Thanks for clarifying what you meant, I just wanted to be sure.

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Questions on metric spaces?

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