Limit of $\frac{3}{n} \sum_{k=1}^n \left[\left(\frac{3k}{n}\right)^2 - \left(\frac{3k}{n}\right)\right]$ using Riemann sums

Question

I am asked to find the following limit:

$$\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left[\left(\frac{3k}{n}\right)^2 - \left(\frac{3k}{n}\right)\right].$$

I am trying to invoke the following rule

$$\int_a^b f(x) dx = \lim_{n \rightarrow \infty } \sum_{k=1}^n f(x+k \Delta x) \Delta x$$

Where $\Delta x = \frac{b-a}{n}$.

So far what I have is that my function $f$ is $f(x)=x^2$ and that $a=1,b=4$. Then I get $$\int_1^4 x^2 dx$$ which is not the correct answer. Can someone help me with where I went wrong? Thanks in advance. Don't necessarily need answer just a hint or someone point out what obvious mistake I am making. Sorry. Also if this is a repeat question I apologize I will take it down, couldn't find one.

Answer 1

By definition of Riemann sum we have

$$\int_{a} ^{b} f(x) \, dx=\lim_{n\to\infty} \frac{b-a} {n} \sum_{k=1}^{n}f\left(a+k\cdot\frac {b-a} {n} \right)$$

which in your case leads to

$$\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left[\left(\frac{3k}{n}\right)^2 - \left(\frac{3k}{n}\right)\right]=\int_{0} ^{3} x^2-x \, dx$$

Refer also to

• Perfect understanding of Riemann Sums

Answer 2

Alternative approach:

Hello from the rectangle-happy Anti-Calculus League.

$$\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left[\left(\frac{3k}{n}\right)^2 - \left(\frac{3k}{n}\right)\right].$$

$\sum_{r=1}^n r^2 = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n.$

$\sum_{r=1}^n r^1 = \frac{1}{2}n^2 + \frac{1}{2}n.$

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$\displaystyle \sum_{k=1}^n (3k)^2 = 9\sum_{k=1}^n k^2 = \frac{9}{3}n^3 + \frac{9}{2}n^2 + \frac{9}{6}n.$

Therefore,
$\displaystyle \sum_{k=1}^{n} \left(\frac{3k}{n}\right)^2 = \frac{1}{n^2} \times \left[\frac{9}{3}n^3 + \frac{9}{2}n^2 + \frac{9}{6}n\right].$

This equals
$\displaystyle \frac{9n}{3} + \frac{9}{2} + \frac{9}{6n}.$

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$\displaystyle \sum_{k=1}^n (3k) = 3\sum_{k=1}^n k = \frac{3}{2}n^2 + \frac{3}{2}n.$

Therefore,
$\displaystyle \sum_{k=1}^{n} \left(\frac{3k}{n}\right) = \frac{1}{n} \times \left[\frac{3}{2}n^2 + \frac{3}{2}n\right].$

This equals
$\displaystyle \frac{3n}{2} + \frac{3}{2}.$

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Therefore, ignoring the leading factor of $\displaystyle \frac{3}{n}$ and ignoring that you are taking a limit, you have the intermediate computation of

$\displaystyle \frac{3n}{2} + 3 + \frac{3}{2n}.$

Applying the leading factor $\displaystyle \frac{3}{n}$ gives

$$\frac{9}{2} + \frac{9}{n} + \frac{9}{2n^2}.\tag1$$

Clearly, when applying the limit, as $n \to \infty$, all but the leftmost term in (1) above go to $0$.

Therefore, the limit is $\displaystyle \left(\frac{9}{2}\right)$.