What is $\lim_{n\to\infty} \left(\sum_{k=1}^n \frac1k\right) / \left(\sum_{k=0}^n \frac1{2k+1}\right)$?

Using Stolz–Cesàro theorem we have:

$$\lim_{n\to\infty} { \sum_{k=1}^{n} {\frac{1}{k}} \over {\sum_{k=0}^{n} {\frac{1}{2k + 1}}}}=\lim_{n\to\infty} { \sum_{k=1}^{n+1} {\frac{1}{k}}-\sum_{k=1}^{n} {\frac{1}{k}} \over {\sum_{k=0}^{n+1} {\frac{1}{2k + 1}}}-{\sum_{k=0}^{n} {\frac{1}{2k + 1}}}}=\lim_{n\to\infty}\frac{2n+3}{n+1}=2$$

Hint Denote the $n$th harmonic number by $$H_n := 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}.$$ Then, the numerator of the given ratio is $H_n$, and the denominator can be written as \begin{align*} 1 + \tfrac{1}{3} + \tfrac{1}{5} + \cdots + \tfrac{1}{2 n + 1} &= \left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots + \tfrac{1}{2 n}\right) - \left(\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{6} + \cdots + \tfrac{1}{2 n}\right) + \tfrac{1}{2 n + 1} \\ &= \left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \cdots + \tfrac{1}{2 n}\right) - \tfrac{1}{2}\left(1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{n}\right) + \tfrac{1}{2 n + 1} \\ &= H_{2 n} - \tfrac{1}{2} H_{n} + \frac{1}{2 n + 1} . \end{align*} Now, using appropriate Riemann sum estimates gives that $$H_n = \log n + O(1).$$

Additional hint So, the denominator is $$\log (2 n) - \tfrac{1}{2} \log n + O(1) = \tfrac{1}{2} \log n + O(1),$$ and so the ratio is $$\frac{\log n}{\tfrac{1}{2} \log n} + O((\log n)^{-1}) = 2 + O((\log n)^{-1}) .$$