# Asymptotic behaviour of a mean

Fix $x>0$ and $c\in\mathbb{N}$. Let $f(x):=\frac{c}{4c-2+2x^2}$ and $$m_N(x):=\frac{1}{N} \sum_{i=0}^{f(x)N} \log(\frac{c N}{2}-i(2c-1))$$

I'm pretty sure $m_N(x)\to\infty$ as $N\to\infty$.

I would like to know the asymptotic behaviour of $m_N(x)$, and I expect to find something like

$$m_N(x)=f(x) \log{N} + g(x) + o(1)\ \ \text{ as }N\to\infty$$

Can you confirm this result? If this is the case, can you help me to compute the constant $g(x)$?

-
Unless you can give some background and context to this question, you are not likely to get any useful answers, and your question is likely to be closed. –  Lee Mosher Jun 11 '12 at 23:28
I'm trying to compute a limit in a Statistical Mechanics model. After some computations, I need to compute that $g(x)$. Just knowing it is true that $m_N(x)=f(x) logN + g(x) + o(1)$ for some $g(x)$ would be a good result. –  user22980 Jun 11 '12 at 23:39
Rather than do the computation, I'll just tell you how I would do it: the sum looks a lot like a Riemann sum. So write the Riemann sum and the integral to which it is an approximation side by side and carefully estimate the differences. –  Anthony Quas Jun 12 '12 at 4:26
I have to guess that $i$ is an integer so, why is the upper limit of the sum the real number $f(x)N$? Is it a mistake or is it so? –  Jon Jun 12 '12 at 7:36
Jon you're right: the sum is over integer $i$. The correct upper bound of the sum would be $$[f(x) N + k(x)]+1$$ where $k(x)=\frac{-x^2}{2c-1+x^2}$ and $[]$ denotes the integer part. But I think it's not important for the asymptotic behaviour of the mean. –  user22980 Jun 12 '12 at 9:52

Let us consider the sum $$m_N(x)=\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{cN}{2}-i(2c-1)\right).$$ The first step to get an asymptotic approximation is to extract the leading term in $N$ to obtain $$m_N(x)=[f(x)]\log N+\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right).$$ When $N$ is finite, we recognize a Riemann series and apply the average theorem. So, there exists a value of argument of the logarithm such that $$m_N(x)=[f(x)]\log N+[f(x)]\log[z(x)].$$ We can take $z(x)=\frac{c}{2}-t$f(x)$(2c-1)$ being $t\in (0,1)$.
Indeed, we can define a partition with $x_i=x_{i-1}+\frac{1}{$f(x)$N}$ and so $$\frac{1}{N}\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-\frac{i}{N}(2c-1)\right)=[f(x)]\Delta x\sum_{i=0}^{[f(x)]N}\log\left(\frac{c}{2}-i$f(x)$(2c-1)\Delta x\right)$$ being $\Delta x=\frac{1}{$f(x)$N}$. But this, in the given limit, is nothing else than $$\int_{\frac{c}{2}}^{\frac{c}{2}-$f(x)$(2c-1)}\log(z)dz<\infty$$ as it should.