Peter Gyarmati

Posted on Tuesday, 22 April, 2003 - 10:29 am:

Peter

David Loeffler

Posted on Tuesday, 22 April, 2003 - 11:10 am:

You show in 2(a) that a_n < something of the order of log(n). So s(n) < something of the order of n log n. Hence 1/s(n) > O(1/(n log n)); and thus

a(n) > O(

n
å
i=2

1/(i logi))

which is divergent (approximately as log log n). This is a very sloppy argument, but I'm sure it can be patched up. It's rather neat that in order to find a lower bound, we first have to find an upper bound! David

Michael Doré

Posted on Tuesday, 22 April, 2003 - 07:53 pm:

An alternative way of looking at it (which doesn't give an explicit rate of divergence unlike David's method) is to note that if a_n didn't tend to infinity then since a_n is strictly increasing there would exist K > 0 such that a_n < K so 1/s_n > 1/(nK).

Hence

a_n = 1 +

n
å
1

1/s_n ® ¥

, which is a contradiction.

Peter Gyarmati

Posted on Tuesday, 22 April, 2003 - 09:31 pm:

Michael,
if I understand you correctly, the following series is divergent:
1+1/2K+1/3K+...+1/nK = 1+1/K(1/2+1/3+...+1/n)
the series in the brackets is divergent (it is the harmonic series), so the whole series is divergent.
Thanks both of you for the help.

Peter