Asymptotics

Oct 2010
3
0
Hello. Here is the problem:

\(\displaystyle f(m) = \sum_{k=2}^{m} \frac{1}{ln k!}\). How to find \(\displaystyle f(m),\) where \(\displaystyle m \rightarrow \inf\).

I need your ideas:) Thanks
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
Well, k! is about (k/e)^k, so 1/log(k) is about 1/(k log k - k). integral 1/k log k up to m is log log m, so that's a reasonable guess. Actually integral 1/(k log k - k) is log(log m - 1) so that 'should be' even more accurate.
 
Similar Math Discussions Math Forum Date
Applied Math
Applied Math