If sequence {x} is in l2, does x_n<k/n follow?

[jpe]

Homework Statement

Suppose we have a sequence {x} = {x_1, x_2, ...} and we know that [itex]\{x\}\in\ell^2[/itex], i.e. [itex]\sum^\infty x^2_n<\infty[/itex]. Does it follow that there exists a K>0 such that [itex]x_n<K/n[/itex] for all n?

Homework Equations

The converse is easy, [itex]\sum 1/n^2 = \pi^2/6[/itex], so there would be a finite upper bound for [itex]\sum^\infty x^2_n<\infty[/itex].

The Attempt at a Solution

I'm stuck. I cannot think of a counterexample and my hunch is that it's true. I was hoping that maybe from [itex]\sum^\infty x^2_n=L[/itex] for some L I could derive a bound for the elements of the sum involving the summation index, but to no avail so far.

 

[jbunniii]

If you had something like [itex]x_n = 1/\sqrt{n}[/itex], there would be no such [itex]K[/itex], right? Of course [itex]\sum x_n^2[/itex] doesn't converge in that case, but can you replace enough of the [itex]x_n[/itex]'s with 0 to achieve convergence?

 

[LCKurtz]

Homework Statement

Suppose we have a sequence {x} = {x_1, x_2, ...} and we know that [itex]\{x\}\in\ell^2[/itex], i.e. [itex]\sum^\infty x^2_n<\infty[/itex]. Does it follow that there exists a K>0 such that [itex]x_n<K/n[/itex] for all n?

Homework Equations

The converse is easy, [itex]\sum 1/n^2 = \pi^2/6[/itex], so there would be a finite upper bound for [itex]\sum^\infty x^2_n<\infty[/itex].

The Attempt at a Solution

I'm stuck. I cannot think of a counterexample and my hunch is that it's true. I was hoping that maybe from [itex]\sum^\infty x^2_n=L[/itex] for some L I could derive a bound for the elements of the sum involving the summation index, but to no avail so far.

When you think about the p series$$\sum_1^\infty \frac 1 {n^p}$$you know that p=1 is the dividing line between convergence and divergence. So as soon as p > 1 even a little, you get convergence. You might be lulled into thinking that anything bigger than n in the denominator will give convergence because of this. But that is wrong because the series$$\sum_2^\infty \frac 1 {n\ln n}$$also diverges. The logarithm grows so slowly that it doesn't help enough to get convergence.

This is a roundabout way of suggesting that if you are looking for counterexamples right on the edge of convergence-divergence, a judicious use of logarithms in the formula might help.

 

[Dick]

I like jbunnii's suggestion better. Make a lot of terms in your series 0.

 

[jpe]

Yeah, that was a good suggestion indeed, it seems almost trivial now. Let {x} be given by
[itex] x_n=1/\sqrt{n}[/itex] if [itex]n=m^2[/itex] for some m, 0 else. So we would have x={1,0,0,1/2,0,0,0,0,1/3,0, ...} for the first few terms. Then:
[itex]\sum_{n=1}^\infty x_n^2 = \sum_{m=1}^\infty (1/\sqrt{m^2})^2=\sum 1/m^2=\pi^2/6[/itex], so [itex]x\in\ell^2[/itex]. However, assume there is a bound [itex]x_n\leq K/n[/itex] for all n. Look at the element with [itex]n=(K+1)^2[/itex]:
[itex]x_{(K+1)^2}=1/(K+1)=(K+1)/(K+1)^2>K/(K+1)^2[/itex]. Thus, K cannot have been a bound. Since K is arbitrary, no such K can exist, and we have found a counterexample!

Thanks for the input guys!