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.

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 $n=3$:
$$
\frac{1+\frac12+\frac13}{1+\frac 13+\frac15+\frac17}= \frac{210+105+70}{210+70+42+30} = \frac{385}{352} \ne 2.
$$
