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

LaTeX Image

exists and equals, say, c, then

LaTeX Image

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

LaTeX Image

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 ® \1-

(1-r)²

¥
å
k=1

æ
è

k
å
l=1

c_l

ö
ø

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 ® 1-
(1-r) ¥
å
k=1
c_k r^k =
lim
r ® 1-
(1-r)² ¥
å
k=1
æ
è k
å
l=1
c_l ö
ø 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 + r² + ¼ and
¥
å
k=1
c_k r^k

, and went from there.

Or were you referring to showing that:

lim
r ® 1-
(1-r)² ¥
å
k=1
æ
è k
å
l=1
c_l ö
ø 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

k
å
l=1

c_l = C_k

. We have by L'Hopital that

lim
r ® 1-

(1-r)

¥
å
k=1

C_k

r^k =

lim
r ® 1-

(1-r)²

¥
å
k=1

C_k r^k.

But (using the notational convenience C₀/0 = 0)

(1-r)

¥
å
k=1

C_k

r^k =

¥
å
k=1

(C_k /k - C_k-1 /(k-1))r^k .

But the last term clearly tends to

¥
å
k=1

(C_k /k - C_k-1 /(k-1)),

which in turn converges to

lim
k ® ¥

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
¥
å
k=1
(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
å
|C_k|/k = ¥

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 ® 1-
(1-r)² ¥
å
k=1
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 e > 0. There exists an integer N such that for all k > N we have |C_k| < ek. Then for any 0 < r < 1:

| ¥
å
k=N+1
C_k r^k |

£

¥
å
k=N+1
ek r^k

£

e/(1-r)²

since

¥
å
k=1
k r^k ] 1/(1-r)².

We can estimate the first N terms of the sum:

| N
å
k=1
C_k r^k |

£

M N
å
k=1
|C_k|r^k

£

N
å
k=1
M k r^k

£

N
å
k=1
M N r^k

£

MN/(1-r)

Putting these together:

| ¥
å
k=1
C_k r^k | £ e/(1-r)² + MN/(1-r)

so

|(1-r)² ¥
å
k=1
C_k r^k | £ e+ MN(1-r) (*)

To recap: we have we have fixed e > 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) < e so:

|(1-r)² ¥
å
k=1
C_k r^k | £ 2 e

therefore
(1-r)² ¥
å
k=1
C_k r^k ® 0

as r ® 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

1/b_n

converges and

Latex image click or follow link to see src

,

then

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).