| Posted on Tuesday, 22 April, 2003 - 10:29 am: |
Peter
| 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
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
| 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 an didn't tend to infinity
then since an is strictly increasing there
would exist K > 0 such that an < K so
1/sn > 1/(nK).
Hence
, which is a contradiction.
| 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