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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.

Martin R
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homosapien
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    How did you come up with $f$ and $a,b$? it seems to me that $f(x)=x^2-x$ and $a=0,b=3$. – Math101 Sep 27 '21 at 19:21
  • for $a$ I used the starting point of the series which is at $k=1$, and I used $\Delta x = \frac{(b-a)}{n}$ to find $b$. I figured its $x^2$ as the variables are being squared. @GBA – homosapien Sep 27 '21 at 19:23
  • wait is it $x^2-x$? I'm more confused lol – homosapien Sep 27 '21 at 19:24
  • @HossienSahebjame Could you please unaccept my current answer? I need to delete it since it is a duplicate of an old one I didn’t remember. You could accept the other very fine answer and refer to the duplicate for the solution by integral. Thanks – user Oct 02 '21 at 07:56

2 Answers2

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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

user
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Alternative approach:

Hello from the rectangle-happy Anti-Calculus League.

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

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

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

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

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

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

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

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 $\displaystyle \left(\frac{9}{2}\right)$.

user2661923
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  • thanks!!! I always appreciate an alternative approach however, for this its time sensitive as I am taking GRE subject soon lol @user2661923 – homosapien Sep 28 '21 at 02:14
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    @HossienSahebjame I elaborated in great detail, just for clarity. In actual fact, my approach, for similar problems, might well be the fastest approach. This is because, with experience, you will recognize, that with each polynomial representation of a summation, you can ignore all but the leftmost term. – user2661923 Sep 28 '21 at 02:23
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    @HossienSahebjame That is, my approach avoids having to waste time carefully associating the summation with a specific integral. If that association takes you longer than $10$ minutes, then my approach will often be fastest. – user2661923 Sep 28 '21 at 02:25