Brad Rodgers

Posted on Monday, 13 January, 2003 - 01:25 am:

Suppose we are given a sequence

a_{n}

such that

$\sum_{n = 1}^{N} a_{n} ~ b N$ ,

for a constant $b$ .

Does this necessarily imply that

$\sum_{n = 1}^{N} a_{n}^{2} ~ c N$ ,

for a constant $c$ ?

Thanks,

Brad

David Loeffler

Posted on Monday, 13 January, 2003 - 09:31 am:

What if we set

a_{2 n} = \sqrt{n}

and

a_{2 n + 1} = 1 - \sqrt{n}

? Then

\sum_{n = 1}^{N} a_{n} ~ \frac{1}{2} N

, but

\sum_{n = 1}^{N} a_{n}^{2} ~ \frac{1}{6} N^{3 / 2}

or something like that.

Maybe the result is true if we constrain $a_{n}$ to be positive real.

David

Michael Doré

Posted on Monday, 13 January, 2003 - 05:10 pm:

It's not. Consider a sequence where for each

n

either

a_{2 n} = a_{2 n + 1} = 1 / 2

a_{2 n} = 3 / 4

a_{2 n + 1} = 1 / 4

. Then

\sum_{n = 1}^{N} a_{n} ~ N / 2

but it should be clear that we make

\sum_{n = 1}^{N} a_{n}^{2} / N

oscillate between 1/4 and 5/16.

Brad Rodgers

Posted on Tuesday, 14 January, 2003 - 02:26 am:

Hrm. I vaguely recall seeing this argument before, or at any rate something very close to it - Michael's that is; in fact, I vaguely recall asking something to the effect of this question before...

Apologies if those memories are accurate, but thanks in either case for the quick responses.

Brad