Re: How to prove an inequality involving sequence of weighted means?

On 02-02-2007 11:06, nicegirl_130@xxxxxxxxx wrote:

I'm not sure if my proof is OK, would like some help please.

Let a_n be a sequence of real numbers, w_n a sequence of positive
weights and s_n = (Sum(i=1,n) (w(i) a(i))/(Sum(i=1, n) w(i)). If
Sum(i=1,oo) w(i) diverges, then

lim inf a_n <= lim inf s_n <= lim sup s_n <= lim sup _a_n (lim inf
and lim sup are quite confusing to me)

The middle inequality holds automatically. My proof for the left one
is as follows (the right one is proved similarly):

If lin inf a_n = -oo , then there's nothing to prove, right?

Right.

So,
suppose lim inf a_n > -oo. For every q < lim inf a(n), there exists
(property of the lim inf) a natural k such that a(n) > q for every n
k. Let S(n) be the sequence of partial sums of w(n) and let w =
infimum{w(1)...w(k)}. Then, for n >k we have s_n = (Sum(i=1, k) w(i)
a(i) + Sum(i= k+1, n) w(i) a(i))/S(n) > (w S(k) + q (S(n) - S(k))/
(S(n) = ((w - q)S(k) + q S(n)) = ((w - q)S(k))/S(n) + q . Putting
y(n) = ((w - q)S(k))/S(n) + q , we get
s(n) > y(n) n for n > k. Therefore, lim inf s(n) >= lim inf y(n)
Since
S(n) diverges and it's terms are positive, S(n) => oo and lim inf
y(n) = lim y(n) = 0 + q = q, because k is fixed and doesn't depend on
n.

Almost. There is a problem when you write that

(Sum(i=1, k) w(i) a(i) + Sum(i= k+1, n) w(i) a(i))/S(n)

is greater than

(w S(k) + q (S(n) - S(k))/S(n).

In this expression you should have Sum(i=1,k)a_i instead of S(k). But
this doesn't affect your argument.

Therefore, lim inf s(n) >= q for every q < lim inf a(n). In order for
this to be true, we must have lim inf s(n) >= lim inf a(n).

These inequalities imply that, if a(n) converges to a, then s(n)
converges to a. This is true even if a = oo or -oo, right?

Almost. The language is not correct, but use "the limit of a(n) is a"
and it will be correct. (If the limit of a sequence is +oo or -oo, it
does not converge.)

Best regards,

Jose Carlos Santos
.

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