I am trying to solve one of the exercises my professor told us to try, just to understand how Integral calculations work using Riemann Sums instead of antiderivatives.

However, I got stuck as I do not know how to continue simplifying...

$\lim_{n \to \infty} \sum_{i=1}^n [(-1+ \frac{4i}{n})^3 + 1]\frac{4}{n}$

$\lim_{n \to \infty} \frac{4}{n}\sum_{i=1}^n [(-1+ \frac{4i}{n})^3 + 1]$

$\lim_{n \to \infty} \frac{4}{n}[\sum_{i=1}^n (-1+ \frac{4i}{n})^3 + \sum_{i=1}^n 1]$

$\lim_{n \to \infty} \frac{4}{n}[\sum_{i=1}^n (-1+ \frac{4i}{n})^3 + n]$

$\lim_{n \to \infty} \frac{4}{n}[(-1+ \frac{4}{n})^3\sum_{i=1}^n (i^3) + n]$

$\lim_{n \to \infty} \frac{4}{n}[(-1+ \frac{4}{n})^3( \frac{n^2(n+1)}{4}) + n]$

$\lim_{n \to \infty} [(\frac{-4}{n}+ \frac{16}{n^3})^3( \frac{n^2(n+1)}{4}) + 4]$

$\lim_{n \to \infty} [(\frac{-4}{n}+ \frac{16}{n^3})^3(\frac{n^4+2n^3+2n^2}{4}) + 4]$

This is where I got stuck. I have no idea what to do with the first parenthesis, especially because of the exponent 3...

Thank you for your help in advance. I am also ready to give explanation on my reasoning or calculations if needed.

  • $\begingroup$ Why do you want to "simplify" that sum? As it is , with the limit, it equals $\;\int_{-1}^3(x^3+1)dx\;$ ... $\endgroup$ – DonAntonio Nov 24 '17 at 16:19
  • $\begingroup$ @donantonio The OP is using Riemann Sums to evaluate the integral. $\endgroup$ – Mark Viola Nov 24 '17 at 16:21


$$\left(-1+\frac{4i}{n} \right)^3=-1+\frac{12i}{n} -\frac{48i^2}{n^2}+\frac{64i^3}{n^3}$$

  • $\begingroup$ Thank you! How did you find thst out? $\endgroup$ – Disjoint Nov 25 '17 at 7:49
  • $\begingroup$ You're welcome. My pleasure. Note $(x+y)^3=x^3+3x^2y+3xy^2+y^3$. $\endgroup$ – Mark Viola Nov 25 '17 at 14:47

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