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

Evaluate
$$ \lim_{n\to\infty}{{1+\frac12+\frac13 +\frac14+\ldots+\frac1n}\over{1+\frac13 +\frac15+\frac17+\ldots+\frac1{2n+1}}} $$

I tried making it into two sums, and tried to make it somehow into an integral, but couldn't find an integral.
The sums I came up with,
$$ \lim_{n\to\infty} { \sum_{k=1}^n {\frac1k} \over {\sum_{k=0}^n {\frac{1}{2k+1}}}} $$

\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*}
