Expressing Express $\lim_{n\to∞} \left(\frac1{n+1} + \frac1{n+2} + ... +\frac1{n+n}\right)$ using Riemann definite integration

Question

Express $\lim_{n\to∞} \left(\frac{1}{n+1} + \frac{1}{n+2} + ... +\frac{1}{n+n}\right)$ as a Riemann integral.

I was able to transform it to $\lim_{n\to ∞}\sum\limits_{r=1}^n \frac1{n+r}$.

I don't know how to go about it from here.

Answer 1

When doing problems like this you want to look for a factor of $\frac{1}{n}$, as this is $\Delta x$. You also want to look for something involving $\frac{k}{n}$, or $\frac{r}{n}$ in your case as you are using $r$ as the index, because these will be the sampling points from the intervals. So I suggest dividing the numerator and denominator by $n$ to get $$\lim _{n \to \infty} \sum_{r=1}^n \frac{1}{n+r} =\lim _{n \to \infty} \sum_{r=1}^n \frac{\frac{1}{n}}{1+\frac{r}{n}}.$$ Now you should be able to figure out the function and the limits of integration. There's actually more than one choice here. You could say $1+\frac{r}{n}$ are the sampling points, and we'll say they're the right endpoints of the intervals (since $r$ starts at $1$ and not $0$). So your sampling points are $x_1=1+\frac{1}{n},x_2=1+\frac{2}{n},\dots,x_n=1+\frac{n}{n}=2$. Thus the integral is from $1$ to $2$. The $r$th term in the sum is $$\frac{\frac{1}{n}}{1+\frac{r}{n}}= \frac{\Delta x}{x_r}= \frac{1}{x_r} \Delta x$$ so in this case your function is $\frac{1}{x}$. Then the limit is equal to the integral $\int_1^2\frac{1}{x}dx$.

Another option is to say that the sampling points are $\frac{r}{n}$, so $x_1 = \frac{1}{n},x_2 = \frac{2}{n}, \dots,x_n = \frac{n}{n}=1$. In this case the integral goes from $0$ to $1$, and the $r$th term of the sum is $$\frac{\frac{1}{n}}{1+\frac{r}{n}} = \frac{\Delta x}{1+x_r} = \frac{1}{1+x_r}\Delta x$$ and so the function is $\frac{1}{1+x}$, and the limit is equal to the integral $\int_0^1 \frac{1}{1+x}dx$. A $u$-substitution should convince you that you get the same answer from both choices.

Answer 2

If $1/n$ is factored out it is $[1/n](1/(1+1/n))+1/(1+2/n)+\cdots +1/(1+n/n),$ which is the right hand Riemann sum with $n$ subdivisions for integrating $1/x$ from $x=1$ to $x=2.$