Hi,

I'm studying the asymptotic behavior $(n \rightarrow \infty)$ of the following formula, where $k$ is a given constant. $$ \frac{1}{n^{k(k+1)/(2n)}(2kn−k(1+k) \ln n)^2}$$

I'm trying to do a series expansion on this equation to give the denominator a simpler form so that it is easier to make an asymptotic analysis.

I used mathematica/wolframalpha to expand the formula (the documents say Taylor expansion is used). http://www.wolframalpha.com/input/?i=1%2F%28n%5E%28k+%28k+%2B+1%29%2F%282+n%29%29+%282+k+n+-+k+%281+%2B+k%29+Log%5Bn%5D%29%5E2%29

However in series expansion at $n \rightarrow \infty$, the result still has $log n$. This is actually a form I prefer, compared to the form $$a_0 + a_1x + a_2 x^2+...$$ Does anyone see how the result may be produced? Any help is much appreciated. Thanks.