Series expansion with remaining $\ln n$

Question

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 $\ln n$. 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.

Answer 1

Hint: $n^{k(k+1)/(2n)}=\mathrm e^{k(k+1)\log(n)/(2n)}=1+u+u^2/2+\ldots$ with $u=k(k+1)\log(n)/(2n)$.

This yields the expansion $$ \frac1{4k^2n^2}\left(1-\frac{(k+1)(k-2)}2\frac{\log n}n+O\left(\left(\frac{\log n}n\right)^2\right)\right). $$

Answer 2

Please be careful, because a Taylor expansion at $\infty$ is not a normal series expansion. For $k>0$ it will surely converge to zero, as $\ln(n)$ grows slower than $n$, and hence the denominator is not bounded, as the logarithm is much slower we can forget him for the asympotics

When we look at $$n^\frac{k(k+1)}{2n}$$ this converges to $1$ because the $n$-th root of $n$ converges to 1. So the asymptotic behaviour should be like \[ \frac{1}{(2k n)^2}=\frac{1}{4k^2n^2}\] which is also the coefficient in the taylor expansion at infinity.