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

1. Feb 17, 2012

### jpe

1. The problem statement, all variables and given/known data
Suppose we have a sequence {x} = {x_1, x_2, ...} and we know that $\{x\}\in\ell^2$, i.e. $\sum^\infty x^2_n<\infty$. Does it follow that there exists a K>0 such that $x_n<K/n$ for all n?

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

3. 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 $\sum^\infty x^2_n=L$ for some L I could derive a bound for the elements of the sum involving the summation index, but to no avail so far.

2. Feb 17, 2012

### jbunniii

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

3. Feb 17, 2012

### LCKurtz

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.

4. Feb 17, 2012

### Dick

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

5. Feb 17, 2012

### jpe

Yeah, that was a good suggestion indeed, it seems almost trivial now. Let {x} be given by
$x_n=1/\sqrt{n}$ if $n=m^2$ 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:
$\sum_{n=1}^\infty x_n^2 = \sum_{m=1}^\infty (1/\sqrt{m^2})^2=\sum 1/m^2=\pi^2/6$, so $x\in\ell^2$. However, assume there is a bound $x_n\leq K/n$ for all n. Look at the element with $n=(K+1)^2$:
$x_{(K+1)^2}=1/(K+1)=(K+1)/(K+1)^2>K/(K+1)^2$. 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!