# 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. 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. 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. 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. 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.