Find the limit of $\lim_{n\to \infty}n^2({1\over{n^3+1^3}}+{1\over{n^3+2^3}}+\cdots+{1\over{n^3+n^3}}).$

Find the limit of $$\lim_{n\to \infty}n^2({1\over{n^3+1^3}}+{1\over{n^3+2^3}}+\cdots+{1\over{n^3+n^3}}).$$

I'm not sure how to evaluate this limit. Any hints or solutions are greatly appreciated. I turned this into $\int_o^1{1\over {1+x^3}}dx$ using algebra. Is this correct or is there a better way?

Your integral is correct, since the terms of your sum can be rewritten as $$\frac{n^2}{n^3+i^3} = \frac{1}{n} \cdot \frac{1}{1+(i/n)^3}$$, which means that your sum is a Riemann sum for the corresponding integral.

If you are having a hard time getting started on evaluating the integral, you can factor $x^3+1 = (x+1)(x^2-x+1)$, and then focus on rewriting your integrand in the following form (using the method of partial fractions) $$\frac{1}{x^3+1} = \frac{A}{x+1} + \frac{B(2x-1)}{x^2-x+1} +\frac{C}{x^2-x+1}$$

(also note that $x^2-x+1 = \left(x-\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$.)

• Why $B(2x-1)$? Also, if the integral converges to some real number does the limit converge to that same real number? – 1233dfv Apr 24 '15 at 6:27
• The $B(2x-1)$ is there to make taking antiderivatives less of a pain. The limit of the Riemann sums will be the value of the integral, by the definition of Riemann integral. – Rolf Hoyer Apr 24 '15 at 6:41

Another method : Thanks to some known properties of digamma function, especially the asymptotic expansion, the limit for $n$ tending to infinity is obtained $=\frac{\ln(2)}{3}+\frac{\pi\sqrt 3}{9}$

Of course, this method is arduous. but it is a nice exercise for the lovers of special functions. 