Re: general ratio test

In article <ekil9q$8it$1@xxxxxxxxxxxxxxxxxxxxxxxxxx>,
"rancid moth" <rancidmoth@xxxxxxxxx> wrote:

[a good amount on a general ratio test]

I had thought about your question before and came up with the
following.

Theorem: Suppose a_n is a positive sequence and a_(n+1)/a_n = 1 -
s/n + o(1/n). (a) If s > 1, then for all t in (1,s) there exists
a C_t > 0 such that a_n < C/n^t for all n, hence sum(n=1,oo) a_n
converges. (b) if s = 1, then no conclusion about the convergence
of sum(n=1,oo) a_n can be drawn. However, if a_(n+1)/a_n = 1 -
1/n + b_n, where sum (b_n)^2 converges, then sum a_n diverges.

Lemma: 1. For x < 1, log(1-x) <= -x.
2. 1/n + 1/(n-1) + ... + 1/N > log(n) - Log(N). Here n > N are
positive integers.

Pf of theorem: (a) Suppose 1 < t < s. The o(1/n) hypothesis
implies there is N such that a_(n+1)/a_n < 1 - t/n, n > N.
Writing a_(n+1)/a_N as a product of ratios, we get

a_(n+1) < a_N(1-t/n)(1-t/(n-1))...(1-t/N).

Take logs and use the Lemma to get

log(a_(n+1)) < log(a_N) + log(1-t/n) + ... + log(1-t/N)

< log(a_N) - t(1/n + ... + 1/N)

< log(a_N) - t(log(n) - log(N)).

Exponentiating, a_(n+1) < a_N*N^t/n^t for n > N, and this proves
(a).

For (b), we can look at sum 1/(nlog(n)) and sum 1/(n*log(n)^2) (n
1). The first series diverges, the second converges,
but in each case a_(n+1)/a_n = 1 - 1/n + o(1/n).

But suppose a_(n+1)/a_n = 1 - 1/n + b_n, where sum (b_n)^2
converges. This time we use log(1-x) = - x + O(x^2) for small x.
Much as above (I'll be brief),

log(a_(n+1)) = log(a_N) + log(1-1/n+b_n) + ... + log(1-1/N+b_N)

= log(a_N) - (1/n + ... + 1/N) (1)

+ O([1/n + b_n]^2) + ... + O([1/N + b_N]^2) (2)

(1) is > log(1/n) + constant. Because sum b_m^2 < oo, sum [1/m +
b_m]^2 < oo, hence (2) > constant. Exponentiate to see a_(n+1) >
C/n for all n, which is strictly stronger than the claim sum a_n
diverges.
.