Re: Infinite series question

jane <jane1806@xxxxxxxxxx> writes:

Suppose, x_i in R, 0 < x_i < 1, sum_i x_i = oo, and

A_n = sum_{i <= i <= n} x_i, so that A_n -> oo.

B_n = sum_{1 <= i <= n} x_i exp[-A_n * xi].

Are there any known results under which conditions on
(x_i), we have that B_n -> 0 as n -> oo ?

I forgot to mention in the previous post, that it is also known that lim
x_i = 0 and i -> oo.


It's true if x_i > c i^(-p) for constants c > 0 and 0 < p < 1/2.
A_n > c int_1^(n+1) t^(-p) dt = c/(1-p) ((n+1)^(1-p) - 1) > b n^(1-p)
for n sufficiently large, if 0 < b < c/(1-p).

Since ln(t)/t is decreasing for large t, for n sufficiently large we have
2 ln(A_n)/A_n < 2 (ln(b) + (1-p) ln(n))/(b n^(1-p)) < c n^(-p) < x_i for
1 <= i <= n.

Then exp(-A_n x_i) <= exp(-2 ln(A_n)) = A_n^(-2) so
B_n <= sum_{i=1}^n x_i A_n^(-2) = A_n^(-1) -> 0 as n -> infinity.

On the other hand, for x_i = i^(-1/2) we have
0 < A_n x_i <= 2 sqrt(2) for i >= n/2, and then
B_n >= A_n exp(-2 sqrt(2)) -> infty.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.

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