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.

Solution 1:

By the Stolz-Cesaro theorem from https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem, $$\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{2n+3}}=2.$$

Solution 2:

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 $$H_n\sim\ln n$$

Solution 3:

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