sum of $\lim \limits_{n \to \infty}\big(\frac{1}{2n}+\frac{1}{2n+1}+\cdots+\frac{1}{3n-2}+\frac{1}{3n-1}\big)$

Question

My own idea is to use Squeeze Rule and use the inequality below to scale $$\ln(n+1)<\sum_{i=1}^{n}\frac{1}{i}<\ln(n)+1$$ But with this method I can't get the same value on both sides.
So what should be the right approach?
Any help will be appreciated!

Answer 1

This is the same as the left Riemann sum for the integral of $1/x$ over the interval $[2,3]$ hence we have $$\begin{align} \lim_{n\to\infty}\left(\frac1{2n}+\frac1{2n+1}+\cdots+\frac1{3n-1}\right) &=\lim_{n\to\infty}\frac1n\sum_{k=0}^{n-1}\frac1{2+\frac{k}n}\\ &=\int_2^3\frac{1}{x}\mathrm{d}x\\ &=\ln{(3)}-\ln{(2)}\\ \end{align}$$

Answer 2

You can use the "atomic" relation that leads to the bounds you cited, $$ \ln(n+1)-\ln(n)\le\frac1n\le\ln(n)-\ln(n-1). $$ Then your sequence has the bounds, telescoping in the upper and lower sums, $$ \ln(3n)-\ln(2n)\le\frac1{2n}+\frac1{2n+1}+\dots+\frac1{3n-1}\le \ln(3n-1)-\ln(2n-1), $$ which then can be used with the "squeeze rule" (or "sandwich lemma").