Calculate: $\lim_n\frac{\ln(2^{\frac{1}{n}})-\ln(n^2)}{1+\frac{1}{2}+...+\frac{1}{n}}$

Question

Find the following limit:

$$\lim_n\frac{\ln(2^{\frac{1}{n}})-\ln(n^2)}{1+\frac{1}{2}+...+\frac{1}{n}}$$

I tried this:

$$\lim_n \frac{\ln(2^{\frac{1}{n}})-\ln(n^2)}{1+\frac{1}{2}+...+\frac{1}{n}}=\frac{\lim_n \ln \frac{2^{\frac{1}{n}}}{n^2}}{\lim_n 1+\frac{1}{2}+...+\frac{1}{n}}=\frac{\ln \lim_n \frac{2^{\frac{1}{n}}}{n^2}}{\lim_n 1+\frac{1}{2}+...+\frac{1}{n}}$$

But then I get $\ln 0$ in the numerator which is undefined. Also I have no idea what to do with the denominator. Any help is appreciated.

Answer 1

Let $H_n = 1+\frac{1}{2}+\ldots+\frac{1}{n}$. It is well known that the limit $$ \lim_{n\rightarrow\infty} (H_n-\ln(n)) $$ exists (it is the Euler-Mascheroni constant $\gamma$, its value is irrelevant for this proof) From the existence of this limit and from the divergence of $H_n$ and $\ln(n)$ we can conclude $$ \lim_{n\rightarrow\infty} \frac{H_n}{\ln(n)} = 1 $$ With this, you get $$ \lim_{n\rightarrow\infty} \frac{\ln\left(2^\frac{1}{n}\right)-\ln\left(n^2\right)}{H_n} = \lim_{n\rightarrow\infty} \frac{\frac{1}{n}\ln(2)-2\ln(n)}{H_n} $$ Dividing numerator and denominator by $\ln(n)$: $$ \ldots = \lim_{n\rightarrow\infty} \frac{\frac{\ln(2)}{n\ln(n)}-2}{\frac{H_n}{\ln(n)}} = \frac{0-2}{1} = -2 $$

Answer 2

With equivalents:

We know the partial sums of the harmonic series $\displaystyle H_n=\sum_{k=1}^n\frac1n\sim_\infty \ln n$. On the other hand $$\ln 2^{1/n}-\ln n^2=\frac 1n\ln 2-2\ln n\sim_\infty -2\ln n,$$ so $$\frac{\ln 2^{1/n}-\ln n^2}{H_n}\sim_\infty\frac{-2\ln n}{\ln n}=-2.$$