A couple convergence questions

[daon]

1) If \(\displaystyle \L \sum a_n\) converges then \(\displaystyle \L \sum a_n^k\) converges.


To be honest, I'm not quite sure what I should apply here. There are many possibilities such as the formal defn of convergence, contractiveness, Cauchy, the many convergence tests, etc. Any ideas?

2) If \(\displaystyle \L \sum a_n\) converges then it does not follow that \(\displaystyle \L \sum \sqrt{a_n}\) converges. Think of an example where this is the case. However, prove that \(\displaystyle \L \sum \frac{\sqrt{a_n}}{n}\) does converge.

The first part here was easy. \(\displaystyle \frac{1}{n^2}\) converges but \(\displaystyle \sum \sqrt{\frac{1}{n^2}} = \sum \frac{1}{n}\) does not.

However on the second part I'm having trouble. There is a hint given saying "can you find a way to apply (a-b)² >= 0?" I have thought about it and come up with:

\(\displaystyle \frac{\sqrt{a_n}}{n} \le \frac{1}{2}(1+\frac{a_n}{n^2})\). I;m not sure if that helps me, but I thought i may have some possibilities since I know \(\displaystyle a_n\) and \(\displaystyle \frac{1}{n^2}\) both converge.

ANy help much appreciated.

-Daon

 

[marcmtlca]

for question 1. doesn't k have to be greater or equal to 1?if this is the case then maybe you could apply the definition of convergence to say that all of the \(\displaystyle |a_n|\)'s past some \(\displaystyle N \in \bf{N}\) are less than 1. it follows that if \(\displaystyle k \geq 1\) then \(\displaystyle |a_n^k| \le |a_n|\) in which case you can squeeze the tail of the sequence by the tail of \(\displaystyle a_n\) to show convergence.

 

[daon]

for question 1. doesn't k have to be greater or equal to 1?if this is the case then maybe you could apply the definition of convergence to say that all of the \(\displaystyle |a_n|\)'s past some \(\displaystyle N \in \bf{N}\) are less than 1. it follows that if \(\displaystyle k \geq 1\) then \(\displaystyle |a_n^k| \le |a_n|\) in which case you can squeeze the tail of the sequence by the tail of \(\displaystyle a_n\) to show convergence.

Yes, you were correct about k, I should have mentioned that. I like your idea.

Well, I know that the limit of a_n must be zero. Thus, \(\displaystyle \forall \epsilon > 0 \,\, \exists N \in \mathbb{N}\) such that \(\displaystyle n > N \Rightarrow |a_n| < \epsilon\)

So, for \(\displaystyle 0 < \epsilon < 1\), \(\displaystyle |a_n| < \epsilon \Rightarrow |a_n^k| < \epsilon\).

I see that the sequence \(\displaystyle a_n^k\) must converge, but I don't see how the series does. Am I missing something?

 

[pka]

Of course #1 is not true!

Consider: \(\displaystyle a_n = \frac{{\left( { - 1} \right)^n }}{{\sqrt n }}\;\& \;k = 2\;\)

 

[daon]

Oops I was a little tired last night...

I didn't read the directions! \(\displaystyle a_n \ge 0\) for all n.

 

[pka]

Well then we know that for some N, \(\displaystyle n \ge N\) we have \(\displaystyle 0 \le a_n < 1\, \Rightarrow \;0 \le \left( {a_n } \right)^k < a_n .\)

 

[daon]

Okay, I get that. Would I then apply the "Tails theorem for series"?

It says that \(\displaystyle \sum a_n\) converges \(\displaystyle \Leftrightarrow\) there is an index m such that \(\displaystyle \sum _{n=m}^{\infty} a_n\) converges.

Can I say let N be the natural number such that \(\displaystyle a_N < 1\). Then take M = max(m,N), where m is the above index. We then know \(\displaystyle 0 \le a_n^k < a_n < 1 \,\, \forall n > M\) and also \(\displaystyle \sum_{n=M}^{\infty}a_n\) converges and is greater then \(\displaystyle \sum_{n=M}^{\infty}a_n^k\). Then, since \(\displaystyle \sum_{n=M}^{\infty}a_n^k\) is bounded and non-decreasing it must converge.

By the Tails theorem for series, \(\displaystyle \sum_{n=M}^{\infty}a_n^k\) converges \(\displaystyle \Leftrightarrow \sum a_n^k\) converges.

I think I had a dream about converging series last night. Scary!

 

[pka]

Yes that will work.
Think about sequesnces of partial sums:
\(\displaystyle S_N = \sum\limits_{n = 1}^N {a_n } \, \Rightarrow \;\left( {S_n } \right) \to S.\)

\(\displaystyle M \ge 1\, \Rightarrow \;\left( {S_{n + M} } \right) \to S\)

 

[daon]

Great, thanks.

Also, I think I've got a hold on the second question, I guess all it takes is a bit of sleep.

From #2, WTS that \(\displaystyle \sum \frac{\sqrt{a_n}}{n}\) converges if \(\displaystyle a_n \ge 0 \forall n\).

Lemma: for all natural numbers k,
\(\displaystyle \L \sum_{n=1}^{k} a_n \,\, \ge \,\, \sum_{n=1}^k \frac{\sqrt{a_n}}{n}\)

If I prove this by induction, then since \(\displaystyle \sum a_n\) is convergent and \(\displaystyle \sum \frac{\sqrt{a_n}}{n}\) is increasing and bounded above by \(\displaystyle \sum a_n\), then it must converge.

Will this work?

Thanks,
-Daon

 

[marcmtlca]

ya a_n needed to be positive, I had assumed that.
I saw a similar problem in my analysis class a few years ago. With the assumption a_n >= 0. my argument works.


Marc

 

[pka]

Lemma: for all natural numbers k,
\(\displaystyle \L \sum_{n=1}^{k} a_n \,\, \ge \,\, \sum_{n=1}^k \frac{\sqrt{a_n}}{n}\)

You cannot show that.
Consider this example: \(\displaystyle a_n = 2^{ - n}\) and compare:
\(\displaystyle \sum\limits_{n = 1}^N {a_n } \quad \& \quad \sum\limits_{n = 1}^N {\frac{{\sqrt {a_n } }}{n}}.\)

The first is less that one the second more than one.

 

[daon]

Okay, ya got me!

I feel all I need to show is that if for n>N, \(\displaystyle n\sqrt{a_n} < 1\) Then I can get that \(\displaystyle \sqrt{a_n}/n < 1/n^2\).

Not sure how I'm going to do that, but I'll try sure enough.

 

[daon]

Actually, I was thinking about applying the Cauchy Condensation test..

\(\displaystyle \L \sum a_n\) converges \(\displaystyle \L \Leftrightarrow \sum 2^na_{2^n}\) converges.

Let b_n be \(\displaystyle \sqrt{a_n}/n\)
Then I get:
\(\displaystyle \L \sum 2^n b_{2^n} = \sum 2^n \frac{\sqrt{a_{2^n}}}{2^n} = \sum \sqrt{a_{2^n}}\)

Does this justify convergence?

Thanks

 

[pka]

\(\displaystyle \L \frac{1}{{n^2 }} + \left( {a_n } \right) \ge \frac{{2\sqrt {a_n } }}{n}\quad \Rightarrow \quad \sum\limits_{k = 1}^\infty {\frac{{2\sqrt {a_n } }}{n}} \le \sum\limits_{n = 1}^\infty {\left[ {\frac{1}{{n^2 }} + \left( {a_n } \right)} \right]}\)

 

[daon]

\(\displaystyle \L \frac{1}{{n^2 }} + \left( {a_n } \right) \ge \frac{{2\sqrt {a_n } }}{n}\quad \Rightarrow \quad \sum\limits_{k = 1}^\infty {\frac{{2\sqrt {a_n } }}{n}} \le \sum\limits_{n = 1}^\infty {\left[ {\frac{1}{{n^2 }} + \left( {a_n } \right)} \right]}\)

Thats so funny, because I just did something similar to that.

I used the fact that \(\displaystyle (\sqrt{a_n}-\frac{1}{2n})^2 > 0\)

To get \(\displaystyle \L \frac{a_n + \frac{1}{4n^2}}{2} \ge \frac{\sqrt{a_n}}{n}\).

But I was thinking.. what right do I have to claim that \(\displaystyle \sum a_n + \frac{1}{4n^2}\) converges? Intuitively it is obvious since they both "terminate".. but how would I show it? Whould breaking the summation appart and saying that \(\displaystyle \sum a_n < C\) and \(\displaystyle \sum \frac{1}{4n^2} < K.\) Thus \(\displaystyle \sum \frac{\sqrt{a_n}}{n} < C+K\)??

 

[pka]

Yes that does work because non-negative series are absolutely sumable.
That means that they can be rearranged.