Limit of $\;\lim\limits_{n\to\infty}[\frac{(\ln(2^{1/n})-\ln(n^2)}{1+1/2+1/3+...+1/n})]$

Question

As the title says the task is to find the limit:$\\$

$$\lim_{n\to\infty}\frac{\ln2^{1/n}-\ln n^2}{1+1/2+1/3+...+1/n})$$

$\\$ I assume I should attempt to somehow reach the form of $\lim_{n\to0}\frac{\ln(1+n)}{n}$ or use the Cesaro-Stolz theorem,but both of those attempts of mine have failed to get me the correct answer. I would appreciate any kind of hints. And also, I'm not allowed to use L'hospital's rule, or any kind of derivation at that.

Thank you in advance!

EDIT: I am not to use integrals either.

Answer 1

It is easy to see that the first fraction limits to zero, it is just $$\frac{\ln 2}{nH_n}.$$ Thus the problem boils down to showing that $$\frac{\ln n}{H_n}\to 1$$ This is either well known or follows from Stolz theorem since $$(n+1)\ln (1+\frac{1}{n})\to 1$$

Answer 2

Use some quick log rules to see that

$$\ln(2^{1/n})-\ln(n^2)=\frac1n\ln(2)-2\ln(n)$$

Also note that

$$\ln(n)=\int_1^n\frac1x\ dx\le\sum_{k=1}^n\frac1k\le1+\int_1^n\frac1x\ dx=1+\ln(n)$$

Thus,

$$\frac{\frac1n\ln(2)-2\ln(n)}{\ln(n)}\le\frac{\ln(2^{1/n})-\ln(n^2)}{\sum_{k=1}^n\frac1k}\le\frac{\frac1n\ln(2)-2\ln(n)}{1+\ln(n)}$$

And apply the squeeze theorem (divide all fractions by $\ln(n)$ if you still need help)