<s>[INST] 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,
$$ \lim_{n\to\infty} { \sum_{k=1}^n {\frac1k} \over {\sum_{k=0}^n {\frac{1}{2k+1}}}} $$
 [/INST] 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
$$
\frac{\ln(n)}{1 + \frac 12 \ln(2n + 1)}
\leq
\frac{1+\frac{1}{2}+\frac{1}{3}+…\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{2n+1}}
\leq 
\frac{1 + \ln(n)}{\frac 12 \ln(2n + 1)}.
$$

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
$$
\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2} \leq \\
1+\frac{1}{3}+\frac{1}{5}+…+\frac{1}{2n+1} \leq \\
1 + \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n+2}\right).
$$
</s>