Express this limit as a definite integral. No interval given. $\lim\limits_{n\to\infty}\sum_{k=1}^n \left(1+\frac{2k}{n}\right)\cdot \frac{2}{n}$

Question

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!

Answer 1

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

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

Answer 2

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 $2\int_0^1 (1+2x)\,dx$.