I'm trying to understand $$\lim_{n\to\infty}\frac{3}{n}\sum_{k=1}^{n} \left(\left(\frac{3k}{n}\right)^2- \left(\frac{3k}{n}\right) \right).$$ I believe we can take $a=0, b=3$ and so this is equivalent to $$\int_{0}^{3} (x^2-x) \, dx.$$ Is this correct?

  • 2
    Yes, this is correct. – szw1710 Sep 20 at 20:27
up vote 4 down vote accepted

If $f$ is Riemann integrable at $[a,b]$ then $$\lim_{n\to +\infty}\frac{b-a}{n}\sum_{k=1}^nf(a+k\frac{b-a}{n})=\int_a^bf(x)dx$$

in your case $$a=0, \; b=3,\; f(x)=x^2-x$$ and $$\int_0^3f=\frac 92$$

Recall that in general by Riemann sum

$$\lim_{n\to \infty}\frac{b-a}n\sum_{k=0}^{n} f\left(a+{k\over n}(b-a)\right)=\int_a^b f(x) dx$$

and in your case by $a=0$, $b=3$ we have

$$\lim_{n\rightarrow\infty}\frac{3}{n}\sum_{k=1}^n\left(\left(\frac{3k}{n}\right)^2-\frac{3k}{n}\right)=\int_0^3(x^2-x) dx$$

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