Hoi,
I was wondering if the following was true. I can't prove it...but it looks true
suppose
for series in
Then
Sounds like something that should be trivially true..but i have hard time showing it..
There is a special case that is true with . Suppose that and . Then
directly by the Chebyshev inequality, and you are done.
I have a feeling that your statement may be false in general but I'm having trouble finding counter-examples.
I am thinking the following: it follows and implying that are Cauchy with limit resp.
Then product as well.... (don't know the proof but product of 2 Cauchy series is something that should be true)
If I'm correct then it should follow that
which is the same as saying that implies
Right? O.o
Actually, here's a proof.
Let and . Then you have that
Next,
Summing over and dividing by yields
The latter two terms converge to 0 by what's given. We use the triangle inequality on the first term
.
This converges to 0 as well, so in the end, the sum on the LHS of the inequality also converges to 0.