Show that absolute sum converges

Hoi,

I was wondering if the following was true. I can't prove it...but it looks true

suppose $\displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n| a_i-p| = \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n |b_i-q| = 0$

for series in $\displaystyle (a_i),(b_i) \subset \mathbb{R}$

Then $\displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^n|a_i b_i -pq| = 0$

Sounds like something that should be trivially true..but i have hard time showing it..

Re: Show that absolute sum converges

Quote:

Originally Posted by

**Dinkydoe** Sounds like something that should be trivially true..but i have hard time showing it..

It is false. Choose for example $\displaystyle a_i=b_i=1/i^2$, then $\displaystyle \sum_{i=1}^{\infty}a_i=\sum_{i=1}^{\infty}b_i=\pi^ 2/6$ but $\displaystyle \sum_{i=1}^{\infty}a_ib_i=\sum_{i=1}^{\infty}1/i^4=\pi^4/90$ and $\displaystyle \pi^2/6\cdot \pi^2/6\neq \pi^2/90$.

P.S. I see that you have changed the hypothesis.

Re: Show that absolute sum converges

yes sorry, I was editing it :/

thanks for your reply though. It looks so convincing to me...and i need it for something. But i cant show it

Re: Show that absolute sum converges

There is a special case that is true with $\displaystyle p,q = 0$. Suppose that $\displaystyle a_1 \leq a_2 \leq \cdots \leq a_n$ and $\displaystyle b_1 \ge b_2 \ge \cdots \ge b_n$. Then

$\displaystyle \frac{1}{n} \sum_{k=1}^n a_k b_k \leq \left(\frac{1}{n} \sum_{k=1}^n a_k\right) \left(\frac{1}{n} \sum_{k=1}^n b_k\right) $

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.

Re: Show that absolute sum converges

I am thinking the following: it follows $\displaystyle |a_i -p| \to 0$ and $\displaystyle |b_i -q|\to 0$ implying that $\displaystyle (a_i),(b_i)$ are Cauchy with limit $\displaystyle p$ resp. $\displaystyle q$

Then product $\displaystyle |a_ib_i-pq|\to 0$ as well.... (don't know the proof but product of 2 Cauchy series $\displaystyle a_n\cdot b_n\to a\cdot b$ is something that should be true)

If I'm correct then it should follow that $\displaystyle 1/n\sum_{i=1}^n|a_ib_i-pq|\to 0$

which is the same as saying that $\displaystyle a_k\to 0$ implies $\displaystyle 1/n \sum_{i=1}^n|a_i|\to 0$

Right? O.o

Re: Show that absolute sum converges

Actually, here's a proof.

Let $\displaystyle a_k = p+ c_k$ and $\displaystyle b_k = q + d_k$. Then you have that

$\displaystyle \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n |c_k| = \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n |d_k| = 0$

Next,

$\displaystyle |a_kb_k-pq| = |c_k d_k+p d_k + q c_k| \leq |c_k d_k|+|p d_k| + |q c_k|$

Summing over $\displaystyle k$ and dividing by $\displaystyle n$ yields

$\displaystyle \frac{1}{n} \sum_{k=1}^n |a_kb_k-pq| \leq \frac{1}{n} \sum_{k=1}^n|c_k d_k|+\frac{1}{n} \sum_{k=1}^n|p d_k| + \frac{1}{n} \sum_{k=1}^n|q c_k|$

The latter two terms converge to 0 by what's given. We use the triangle inequality on the first term

$\displaystyle \left(\frac{1}{n} \sum_{k=1}^n|c_k d_k|\right)^2 \leq \left(\frac{1}{n} \sum_{k=1}^n| c_k|\right)\left(\frac{1}{n} \sum_{k=1}^n|d_k|\right) $.

This converges to 0 as well, so in the end, the sum on the LHS of the inequality also converges to 0.

Re: Show that absolute sum converges

haha you genious. Thank you kindly :)

Re: Show that absolute sum converges

Haha, no problem. It's a cool question though