I am having trouble trying to convert a limit to a definite integral. I am unsure about how to go about this. I have already tried googling this but can not find anything that is comprehensive enough for me to learn from.

Here's the limit:

$$\lim_{n\rightarrow \infty}\sum_{k=1}^n \left(1+\frac{2k}{n}\right)\cdot \frac{2}{n}$$

I need to express this as a definite integral but cannot figure out how. My textbook is not clear and doesn't include an example, and my professor did not explain this.

Thank you!

  • $\begingroup$ It's the Riemann sum for $2\int_0^1 (1+2x)\,dx$. $\endgroup$ – Mark Viola Jul 29 '17 at 3:33
  • $\begingroup$ Thank you mark, at least I know if I have it right. Would you be able to write a detailed explanation? I'm not entirely sure how you got that. $\endgroup$ – JustHeavy Jul 29 '17 at 3:35
  • $\begingroup$ You're welcome. I've posted a solution which provides the coveted details. $\endgroup$ – Mark Viola Jul 29 '17 at 3:49

The goal is to represent the limit

$$\lim_{n\rightarrow \infty}\sum_{k=1}^n \left(1+\frac{2k}{n}\right)\cdot \frac{2}{n}$$

as an integral.

In fact, any integral like $\int_a^b f(x) dx$ can be approximated as a sum of $n$ rectangles:

$$\int_a^b f(x)dx \approx \sum_{k=1}^n f(a + k\cdot\Delta x)\cdot \Delta x$$

A picture shows why— here, $\Delta x$ is the width of the rectangles (it's equal to the length of the interval divided into $n$ equal pieces), $(a+k\Delta x)$ is the x-position of the $k$th rectangle, and $f(a+k\cdot \Delta x)$ is its height so that the left tip of the rectangle touches the curve $f(x)$.

If we increase the number of rectangles $n$, the sum should become a more and more accurate approximation of the integral. Eventually, if the limit exists, the approximation will become exact:

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

If we match this general pattern against the equation you're given, it looks like:

  • $\Delta x \longleftrightarrow \frac{2}{n}$ is the rectangle width.
  • $f(a + k \Delta x) \longleftrightarrow (1 + k\cdot \frac{2}{n})$
  • So the left endpoint $a$ is 1.
  • And $f(x) = x$, nothing more complicated.
  • And we can solve for the right endpoint $b$ because we know that $\Delta x \equiv \frac{b-a}{n}$ by definition of these equally-spaced rectangles and $a = 1$ as we have found. So $\frac{2}{n} = \frac{b-1}{n}$ so $b=3$.

We now have all of our components and can write

$$\lim_{n\rightarrow \infty}\sum_{k=1}^n \left(1+ \frac{2k}{n}\right)\cdot \frac{2}{n} = \fbox{$\int_{1}^3 x \, dx$}$$

  • $\begingroup$ Thank you! This is exactly what I was looking for. It is even the answer I got myself, I just didn't have a way to verify it. $\endgroup$ – JustHeavy Jul 29 '17 at 3:53
  • $\begingroup$ @DevHeavy Glad it helped! :) $\endgroup$ – user326210 Jul 29 '17 at 3:58
  • $\begingroup$ Now that I think about it, why does f(x) = x ? The only reason I knew that is because I had it memorized. Thank you for your time. $\endgroup$ – JustHeavy Jul 29 '17 at 3:58
  • $\begingroup$ If your problem instead said $\lim_{n\rightarrow \infty}\sum_{k=1}^n \sin^2\left(1+\frac{2k}{n}\right) \cdot \frac{2}{n}$, the function to be integrated would have been $f(x) = \sin^2(x)$. Does that example help? $\endgroup$ – user326210 Jul 29 '17 at 4:00
  • $\begingroup$ Not exactly, I guess I don't understand why (1+2k/n) = x am I supposed to always just assume that the input of a function will always = x? Is it too hard to prove? $\endgroup$ – JustHeavy Jul 29 '17 at 4:05

If the Riemann integral $\int_0^1 f(x)\,dx$ exists, then it can be written as the limit

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

Using $(1)$ with $f(x)=2(1+2x)$, $a=0$ and $b=1$ reveals that

$$\int_0^1 2(1+2x)\,dx=\lim_{n\to \infty}\sum_{k=1}^n 2\left(1+\frac {2k}n\right)\,\frac1n=\lim_{n\to \infty}\sum_{k=1}^n \left(1+\frac {2k}n\right)\,\frac2n$$

Therefore, the limit of interest is simply the Riemann Sum of the integral $2\int_0^1 (1+2x)\,dx$.

  • $\begingroup$ Thank you again, mark. I really appreciate the time you put into answering my question, but I'm going to go with user326210's answer since it more completely answers my question. $\endgroup$ – JustHeavy Jul 29 '17 at 3:53
  • $\begingroup$ Well, you can up vote multiple answers. $\endgroup$ – Mark Viola Jul 29 '17 at 3:57
  • $\begingroup$ There is a mistake above $\int_{0}^{1} 2(1+2x)\,dx \neq \lim \limits_{n\to \infty}\sum \limits_{k=1}^n 2\left(1+\frac kn\right)\,\frac 1 n$ $\endgroup$ – john Aug 12 '18 at 6:51
  • $\begingroup$ @john Thank you for alerting me to the typographical error. I have edited it accordingly. $\endgroup$ – Mark Viola Aug 12 '18 at 18:06

I would like to correct the answer above/below me, with my own comments.

If the Riemann integral $\int_a^b f(x)\,dx$ exists, then it can be written as the limit of a special sum known as a Riemann sum

$$\int_a^b f(x)\,dx=\lim_{n\to \infty}\sum_{k=1}^n f(c_k) \Delta x \tag 1 $$ where $c_k = a + \frac{b-a}{n} \cdot k $ and $ \Delta x = \frac{b-a}{n}$. The formula for $c_k$ are right endpoints of each of the n uniform width subintervals.

Note that the choice of $c_k$ is not unique and different $c_k$ will produce different functions with different limits for the integral. However the final value for the definite integral should end up being the same.

I will choose $c_k= 0 + \frac{1-0}{n} \cdot k = \frac{k}{n} $ which forces $ \Delta x = \frac{1-0}{n} = \frac 1 n $.

It may seem more natural to choose $c_k= 1 + \frac{3-1}{n} ~ k$ and $ \Delta x = \frac 2 n $ . This will lead to the first answer posted above. I will leave it to you to read that answer.

Using some algebra we can rewrite the original Riemann sum in the appropriate 'integral ready' form using our choice $c_k= \frac{k}{n}$ and $ \Delta x =\frac 1 n $:

$$\begin{align}\lim_{n\rightarrow \infty}\sum_{k=1}^n \left(1+\frac{2k}{n}\right)\cdot \frac{2}{n} &= \lim_{n\rightarrow \infty}\sum_{k=1}^n 2\left(1+ 2 \cdot \frac{k}{n}\right)\cdot \frac{1}{n} \\ &=\lim_{n\rightarrow \infty}\sum_{k=1}^n 2\left(1+ 2 \cdot c_k\right)\cdot \Delta x \\ &= \lim_{n\rightarrow \infty}\sum_{k=1}^n f(c_k) \cdot \Delta x \\ &= \int_{0}^{1} f(x) ~dx \end{align} $$

Notice how the $c_k$ becomes the $x$ in the definite integral. It follows the Riemann Sum is equal to the integral $\int_0^1 2(1+2x)\,dx$.


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