TITLE: Evaluating some Limits as Riemann sums.
QUESTION [13 upvotes]: I really have difficulties with Riemann Sums, especially the ones as below:
$$\lim_{n\to\infty} \left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n}\right)$$
When i try to write this as a sum, it becomes $$\frac { 1 }{ n } \sum _{ k=1 }^{ 2n } \frac { 1 }{ 1+\frac { k }{ n }  } .$$   The problem is, however, to be able to compute this limit as an integral I need to have this sum from $1$ to $n$. There are some other questions like this, but if I can understand it, I will be able solve others.

REPLY [2 votes]: We can rewrite our expression as 
$$\frac{1}{2n}\sum_{k=1}^{2n}\frac{2n}{n+k}$$
and then as 
$$\frac{1}{2n}\sum_{k=1}^{2n}\frac{1}{\frac{1}{2}+\frac{k}{2n}}.$$
This is a Riemann sum for
$$\int_0^1 \frac{1}{\frac{1}{2}+x}\,dx.$$
The integral, and therefore the desired limit, is $\ln(3/2)-\ln(1/2)$, or more simply $\ln 3$.