I'm working through a statistical physics book and one of the problems makes the claim that the quantity:
$$H =\frac{1}{N}\sum_{n=1}^N\frac{1}{\frac{a}{N} + \frac{b}{N^{5/3}}(n^{5/3}-(n-1)^{5/3})}$$ can be expressed in the large $N$ limit as the integral: $$H = N\int_0^1 \frac{\text{d}u}{a+\frac{5}{3}b u^{2/3}}.$$
However, I am unable to arrive at that result. I've tried to use LRAM but I got a different result. Any tips will be appreciated.
Observe that, for $n$ large, $$ n^{5/3} - (n-1)^{5/3} = n^{5/3} \left[1- \left(1 - \frac{1}{n}\right)^{5/3}\right] \sim n^{5/3} \frac{5}{3n} = \frac{5}{3} n^{2/3} $$ Using the above approximation (even if, in my opinion, in this case it is not fully justifiable), you can write $H$ as $$ H \sim N \cdot \sum_{n=1}^N \frac{1}{N} \cdot \frac{1}{a + \frac{5}{3} b \left(\frac{n}{N}\right)^{2/3}}\,. $$ Finally, you can use the fact that, if $f$ is a continuous function, then $$ \sum_{n=1}^N \frac{1}{N} f\left(\frac{n}{N}\right) \to \int_0^1 f(u)\, du, $$ as $N\to +\infty$.