Proof for $\log\frac{2n+1}{n+1}<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}<\log 2$

We have $$\frac{1}{n+k}>\log\left(1+\frac{1}{n+k}\right) $$ hence: $$\sum_{k=1}^{n}\frac{1}{n+k}>\log\prod_{k=1}^{n}\frac{n+k+1}{n+k}=\log\frac{2n+1}{n+1}$$ while: $$\frac{1}{n+k}=\frac{\frac{1}{n+k-1}}{1+\frac{1}{n+k-1}}<\log\left(1+\frac{1}{n+k-1}\right)$$ gives: $$\sum_{k=1}^{n}\frac{1}{n+k}<\log\prod_{k=1}^{n}\frac{n+k}{n+k-1}=\log2$$ as wanted.

Riemann sums do this pretty quickly. $\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$ is a lower Riemann sum for $\int_n^{2n} {1 \over x}\,dx = \log 2$, and an upper Riemann sum for $\int_{n+1}^{2n + 1} {1 \over x}\,dx = \log {2n +1 \over n + 1}$, so one has $$\log {2n +1 \over n + 1} < \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n} < \log 2 $$

As shown in $(1)$ here, $$ \begin{align} H(2n)-H(n) &=\sum_{k=n+1}^{2n}\frac1k\\ &=\sum_{k=1}^{2n}\color{#C00000}{\frac1k}-\sum_{k=1}^n\color{#00A000}{\frac1k}\\ &=\sum_{k=1}^n\left(\color{#C00000}{\frac1{2k-1}+\frac1{2k}}-\color{#00A000}{\frac1k}\right)\\ &=\sum_{k=1}^n\left(\frac1{2k-1}-\frac1{2k}\right)\\ &=\sum_{k=1}^{2n}(-1)^{k-1}\frac1k\tag{1} \end{align}$$ which is a series of positive terms, $\left(\frac1{2k-1}-\frac1{2k}\right)$, whose limit is $\log(2)=\sum\limits_{k=1}^\infty(-1)^{k-1}\frac1k$. Thus, each partial sum is less than $\log(2)$. That is, $$ \sum_{k=n+1}^{2n}\frac1k\lt\log(2)\tag{2} $$ Furthermore, $$ \begin{align} e^{\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}} &\gt\left(1+\frac1{n+1}\right)\left(1+\frac1{n+2}\right)\cdots\left(1+\frac1{2n}\right)\\ &=\frac{2n+1}{n+1}\tag{3} \end{align} $$ Taking the log of $(3)$, we get $$ \sum_{k=n+1}^{2n}\frac1k\gt\log\left(\frac{2n+1}{n+1}\right)\tag{4} $$ Putting together $(2)$ and $(4)$ yields $$ \log\left(\frac{2n+1}{n+1}\right)\lt\sum_{k=n+1}^{2n}\frac1k\lt\log(2)\tag{5} $$