Evaluating some Limits as Riemann sums.

Question

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.

Answer 1

$$ \frac{1}{n} \sum_{k=1}^{2n} \frac{1}{1+\frac{k}{n}}= \sum f(x_k)\,\Delta x. $$ where $f(x) = \dfrac{1}{1+x}$ and $\Delta x = \dfrac 1 n$. The variable goes from $1/n\to0$ to $(2n)/n=2$. Hence the sum approaches $$ \int_0^2 \frac{1}{1+x}\,dx. $$

Answer 2

With Eulero-Mascheroni : $$\sum_{k = 1}^{n}\frac{1}{k} - \log{n} \rightarrow \gamma$$ $$\sum_{k = 1}^{3n}\frac{1}{k} - \log{3n} \rightarrow \gamma$$ so $$\sum_{k = n+1}^{3n}\frac{1}{k} - \log{3n} +\log{n} \rightarrow 0$$ and then$$\sum_{k = n+1}^{3n}\frac{1}{k} \rightarrow \log{3}$$

Answer 3

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 $\ln 3$.