Brad Rodgers

Posted on Tuesday, 15 July, 2003 - 11:43 pm:

Hi. How does one directly prove the theorem that: for a sequence c_n , if

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exists and equals, say, c, then

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It seems to me that the most motivated way to go about this is to show that the limits can be interchanged in

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but I can't seem to be able to do this without a complete mess.

The theorem is known; as the title alludes to, it was apparently first proven by Frobenius. But all the proofs I've come across so far prove a far more general version and require a good deal of apparatus...

Brad

Michael Doré

Posted on Wednesday, 16 July, 2003 - 01:44 am:

Hint: Instead try writing the limit:

lim_{r \to \ 1 -} (1 - r)^{2} \sum_{k = 1}^{\infty} (\sum_{l = 1}^{k} c_{l}) r^{k} .

Brad Rodgers

Posted on Wednesday, 16 July, 2003 - 02:46 am:

Very clever. Just out of curiosity, did you finish by evaluating a telescopic sum? That's the method I saw, but it's probably not the only way...

Thanks,

Brad

Michael Doré

Posted on Wednesday, 16 July, 2003 - 12:59 pm:

Hi Brad,

Which part are you referring to? Are you referring to proving:

$lim_{r \to 1 -} (1 - r) \sum_{k = 1}^{\infty} c_{k} r^{k} = lim_{r \to 1 -} (1 - r)^{2} \sum_{k = 1}^{\infty} (\sum_{l = 1}^{k} c_{l}) r^{k}$

I guess you could do this by using a telescoping sum - I actually did it by multiplying the absolutely convergent series $1 + r + r^{2} + \dots$ and $\sum_{k = 1}^{\infty} c_{k} r^{k}$ , and went from there.

Or were you referring to showing that:

$lim_{r \to 1 -} (1 - r)^{2} \sum_{k = 1}^{\infty} (\sum_{l = 1}^{k} c_{l}) r^{k} = c ?$

I don't immediately see how you could do this with a telescoping sum - but it follows quickly by an epsilon-delta argument (available on request!). If you have a proof of this based on a telescoping sum I'd be interested to see it.

Michael

Brad Rodgers

Posted on Wednesday, 16 July, 2003 - 08:16 pm:

It was in proving the second statement that I used a telescopic sum. First, let

\sum_{l = 1}^{k} c_{l} = C_{k}

. We have by L'Hopital that

lim_{r \to 1 -} (1 - r) \sum_{k = 1}^{\infty} \frac{C_{k}}{k} r^{k} = lim_{r \to 1 -} (1 - r)^{2} \sum_{k = 1}^{\infty} C_{k} r^{k} .

But (using the notational convenience

C_{0} / 0 = 0

)

(1 - r) \sum_{k = 1}^{\infty} \frac{C_{k}}{k} r^{k} = \sum_{k = 1}^{\infty} (C_{k} / k - C_{k - 1} / (k - 1)) r^{k} .

But the last term clearly tends to

\sum_{k = 1}^{\infty} (C_{k} / k - C_{k - 1} / (k - 1)),

which in turn converges to

lim_{k \to \infty} C_{k} / k = c

. I'm also interested in seeing your method.

Brad

Michael Doré

Posted on Wednesday, 16 July, 2003 - 10:50 pm:

Hi Brad,

It's a nice approach. The only thing that worries me is the application of L'Hopital. There doesn't seem to be any reason that $\sum_{k = 1}^{\infty} (C_{k} / k) r^{k}$ should be differentiable at r = 1. The radius of convergence of the power series is certainly no more than 1, so we can differentiate it term by term at any - r - < 1, but it is possible that $\sum | C_{k} | / k = \infty$ in which case I can't presently see how to justify the application of L'H.

The way I had in mind was:

WLOG we can assume c = 0 (otherwise just replace $c_{k}$ with $c_{k} - c$ in the conclusion). Then $C_{k} = o (k)$ and we just need to show:

$lim_{r \to 1 -} (1 - r)^{2} \sum_{k = 1}^{\infty} C_{k} r^{k} = 0$

Note that as $C_{k} = o (k)$ there exists M > 0 such that $C_{k} < Mk$ for all k. Now let $ε > 0$ . There exists an integer N such that for all k > N we have $| C_{k} | < ε k$ . Then for any 0 < r < 1:

$\begin{matrix} | \sum_{k = N + 1}^{\infty} C_{k} r^{k} | & \leq & \sum_{k = N + 1}^{\infty} ε k r^{k} \\ \leq & ε / (1 - r)^{2} \end{matrix}$

since

$\sum_{k = 1}^{\infty} k r^{k}] 1 / (1 - r)^{2} .$

We can estimate the first N terms of the sum:

$\begin{matrix} | \sum_{k = 1}^{N} C_{k} r^{k} | & \leq & M \sum_{k = 1}^{N} | C_{k} | r^{k} \\ \leq & \sum_{k = 1}^{N} M k r^{k} \\ \leq & \sum_{k = 1}^{N} M N r^{k} \\ \leq & MN / (1 - r) \end{matrix}$

Putting these together:

$| \sum_{k = 1}^{\infty} C_{k} r^{k} | \leq ε / (1 - r)^{2} + MN / (1 - r)$

so

$| (1 - r)^{2} \sum_{k = 1}^{\infty} C_{k} r^{k} | \leq ε + MN (1 - r) (*)$

To recap: we have we have fixed $ε > 0$ and proved there exists M, N > 0 such that (*) holds for all 0 < r < 1. Then if r is sufficiently close to 1 then $MN (1 - r) < ε$ so:

$| (1 - r)^{2} \sum_{k = 1}^{\infty} C_{k} r^{k} | \leq 2 ε$

therefore $(1 - r)^{2} \sum_{k = 1}^{\infty} C_{k} r^{k} \to 0$ as $r \to 1 -$ as required. I wonder if any of this can be used to prove your interesting conjectures about Cesaro limits (from about a year ago) that stumped everyone on the board who tried it!

Brad Rodgers

Posted on Friday, 18 July, 2003 - 08:02 pm:

Are you talking about this thread ? If so, I can't yet see an application, but there very well could be one. I'll think about it. More generally, I think it's reasonable to conjecture that if

\sum 1 / b_{n}

converges and

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,

then

\sum a_{n}

converges as well. A condition that b_n > 0 may have to be added -- I haven't given it enough thought to know for sure. I can't find a counter-example in either case.

Did you have another thread in mind?

Thanks for the proof,

Brad

Michael Doré

Posted on Friday, 18 July, 2003 - 08:50 pm:

Yes that was indeed the thread I was thinking of (or what's left of it). I don't really see how to apply this either - in fact when I made that comment I couldn't remember exactly what your conjecture was except that it involved Cesaro limits and that no-one could do it. (Maybe sometime we should compile all of your unsolved NRICH questions into a single thread!)

For your latest conjecture I think you're going to need some extra condition on a_n (perhaps say a_n is decreasing?). At the moment it's quite easy to find counter-examples (I won't say exactly how, since you haven't had time to think about it yet).