Limit of sequence - hard one How to compute $\displaystyle\lim_{n \to \infty}{\sum_{k=1}^{n}{\frac{k^2n^3}{k^6-4n^6}}}$? Any hints?
 A: If you divide the top and bottom by $n^6$, you can write everything as $\frac{1}{n}$ times a function of $\frac{k}{n}$. Then the stated sum becomes the Riemann sum of the integral of that function over the interval $(0,1)$. To evaluate the sum it's then enough to calculate the integral. 
To integrate that function is a separate problem. The function is $f(x)=\frac{x^2}{x^6 - 4}$. The denominator factors into $(x^3-2)(x^3+2)$, so this suggests using the method of partial fractions. Once you write $f(x)$ as a sum of two functions, with denominators $x^3-2$ and $x^3+2$, then integrating each of those functions is straightforward.
A: Try to transform it into a Riemann sum:
$$ \sum_{k=1}^n \frac{k^2n^3}{k^6-4n^6} = \frac{1}{n} \sum_{k=1}^n \frac{\frac{k^2}{n^2}}{\frac{k^6}{n^6}-4}$$
As $n$ goes to infinity, the limit is $\displaystyle \int_0^1 \frac{x^2}{x^6-4}dx$
A: As pointed by the other answers, a Riemann sum argument leads to the integral
$$\int_{0}^{1}\frac{x^2}{x^6-4}\,dx = \frac{1}{3}\int_{0}^{1}\frac{dy}{y^2-4}=\color{red}{-\frac{\log 3}{12}}\tag{1}$$
that we have just computed by using the substitution $x^3=y$ and partial fraction decomposition.
