Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute:  $\displaystyle\lim_{n\rightarrow \infty  }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\displaystyle\lim_{n\rightarrow \infty  }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\rightarrow \infty  }\frac{2s}{s}=2$.
Is that ok?
Thanks.
 A: By the Stolz-Cesaro theorem from https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem,
$$\lim_{n\rightarrow \infty  }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{2n+3}}=2.$$
A: Hint
The numerator is $H_n$ and the denominator is $H_{2n+1}-\frac12H_n$.
Also,
$$\frac{H_n}{H_{2n+1}-\frac12H_n}=\frac1{-\frac12+\frac{H_{2n+1}}{H_n}}$$
and
$$H_n\sim\ln n$$
A: If the numerator is always twice the denominator, then this works provided you include a proof of that.  Let's try it when $n=2$:
$$
\frac{1+\frac12}{1+\frac13+\frac15} = \frac{30+15}{30+10+6}= \frac{45}{46}\ne 2.
$$
Let's try it when $n=3$:
$$
\frac{1+\frac12+\frac13}{1+\frac 13+\frac15+\frac17}= \frac{210+105+70}{210+70+42+30} = \frac{385}{352} \ne 2.
$$
A: I don't know if there's still any interest in this question from more than a year ago, but here's an elementary solution:
The numerator in the problem is $H_n=1+\frac12+\frac13+\dots+\frac{1}{n},$ the $n^{\text{th}}$ harmonic number.
The denominator in the problem is
\begin{align}
1+\frac13+\frac15+\dots+\frac1{2n+1}&=H_{2n+1}-\big(\frac12+\frac14+\frac16+\dots+\frac{1}{2n}\big)
\\&=H_{2n+1}-\frac12 \big(1+\frac12+\frac13+\dots+\frac{1}{n}\big)
\\&=H_{2n+1}-\frac12 H_n.
\end{align}
Next note that
\begin{align}
H_{2n+1}-H_n&=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\dots+\frac{1}{2n+1}
\\&\le\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+1}+\dots+\frac{1}{n+1} \scriptsize{\quad(n+1\text{ terms})}
\\&=1.
\end{align}
Of course, $H_{2n+1}-H_n$ is always positive, being a sum of one or more positive fractions, so we have
$$0\lt H_{2n+1}-H_n \le 1$$
for all non-negative integers $n.$
Since $H_n\to\infty$ as $n\to\infty,$ it follows that
$$
\lim_{n\to\infty}\frac{H_{2n+1}-H_n}{H_n}=0.
$$
We can now solve the original problem:
\begin{align}
\displaystyle\lim_{n\rightarrow \infty  }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}&=\lim_{n\to\infty}\frac{H_n}{H_{2n+1}-\frac12 H_n}
\\&=\lim_{n\to\infty}\frac{H_n}{\frac12 H_n+(H_{2n+1}-H_n)}
\\&=\lim_{n\to\infty}\frac{1}{\frac12+\frac{H_{2n+1}-H_n}{H_n}}
\\&=\frac{1}{\frac12+\lim_{n\to\infty}\frac{H_{2n+1}-H_n}{H_n}}
\\&=\frac{1}{\frac12+0}
\\&=2.
\end{align}
A: With the Euler-MacLaurin formula, we have
numerator = $\ln(n)+O(1)$ and denominator = $\frac12 \ln(2n+1) + O(1)$, so the limit is
$\lim_{n \to \infty}\frac{\ln(n)+O(1)}{\frac12 \ln(2n+1) + O(1)}= 2.$
