Calculating $\lim _{n\to \infty \:}\left(\frac{1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n+1}}\right)$

By Stolz-Cesaro we have: $\lim\limits_{n \to \infty} \frac{1 + \frac12 + \dots + \frac1n}{1 + \frac13 + \frac15 + \dots + \frac1{2n+1}} = \lim\limits_{n \to \infty} \frac{\frac1{n+1}}{\frac1{2n + 3}} = \lim\limits_{n \to \infty} \frac{2n+3}{n+1} = 2$

One approach is as follows: it suffices to note that $$ \sum_{k=2}^n\frac{1}{k} \leq \int_1^n \frac 1x \,dx \leq \sum_{k=1}^n\frac{1}{k}, \\ \sum_{k=2}^{n+1}\frac{1}{2k-1} \leq \int_1^{n+1} \frac 1{2x-1} \,dx \leq \sum_{k=1}^{n+1}\frac{1}{2k-1}, $$ and apply the squeeze theorem. In particular, we can use the above to get $$ \frac{\ln(n)}{1 + \frac 12 \ln(2n + 1)} \leq \frac{1+\frac{1}{2}+\frac{1}{3}+…\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{2n+1}} \leq \frac{1 + \ln(n)}{\frac 12 \ln(2n + 1)}. $$

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Another approach: note that adding a final $\frac 1{2n + 2}$ to $1/2$ times the numerator yields $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2}$, and $$ \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2} \leq \\ 1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{2n+1} \leq \\ 1 + \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2}\right). $$

\begin{align*} \text{Required limit}&=\lim_{n\to\infty}\left(\dfrac{\sum\limits_{i=1}^n\dfrac1i}{\sum\limits_{i=1}^{2n+1}\dfrac1i-\sum\limits_{i=1}^n\dfrac1{2i}}\right)\\ &=\lim_{n\to\infty}\left(\dfrac{H_n}{H_{2n+1}-\dfrac12H_n}\right)\\ &=\lim_{n\to\infty}\left(\dfrac{\dfrac{H_n}{\log(2n+1)}}{\quad\dfrac{H_{2n+1}}{\log(2n+1)}-\dfrac{H_n}{2\log(2n+1)}\quad}\right)\\ &=\dfrac{1}{1-\dfrac12}\\ &=\boxed2 \end{align*}