# Series expansion with remaining $log n$

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.

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It seems that it is the $n^{k(k +1)/(2n)}$? – ELW May 17 '13 at 9:19
the series expansion is easiest if you first take the logarithm, and then you find directly a powerseries in $n^{-1}\ln n$, $$-\frac{k(k+1)}{2n}\ln n+2\ln(2kn)\left[\sum_{p=0}^{\infty}\frac{1}{p}k^p(1+k)^p(2kn)^{-p}(\ln n)^p \right]$$ – Carlo Beenakker May 17 '13 at 12:47