Limit of sequence - hard one

Question

How to compute $\displaystyle\lim_{n \to \infty}{\sum_{k=1}^{n}{\frac{k^2n^3}{k^6-4n^6}}}$? Any hints?

score 5 · Answer 1 · answered May 10 '15 at 20:18

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.

score 3 · Answer 2 · answered May 10 '15 at 20:19

3

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 $\displaystyle \int_0^1 \frac{x^2}{x^6-4}dx$

answered May 10 '15 at 20:19

Beni Bogosel

23,381

n^3 / n^6 = n^2? are you sure? – hildred May 11 '15 at 05:01
I have put a factor $1/n$ before the sum. – Beni Bogosel May 11 '15 at 06:21

score 2 · Answer 3 · answered May 10 '15 at 20:34

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.

Limit of sequence - hard one

3 Answers3