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.

With the Euler-MacLaurin formula, we have
numerator = $\ln(n)+O(1)$ and denominator = $\frac12 \ln(2n+1) + O(1)$, so the limit is
$\lim_{n \to \infty}\frac{\ln(n)+O(1)}{\frac12 \ln(2n+1) + O(1)}= 2.$
