MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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).

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.

share|cite|improve this question
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.