$ \lim_{n\to\infty} \frac{1}{(\ln n)^2} \sum_{i=1}^n\ln( i^{1/i}) $ =?

Question

What is the value of the following limit? $$ \lim_{n\to\infty} \frac{1}{(\ln n)^2} \sum_{i=1}^n \ln (i^{1/i}) .$$

Answer 1

I'm arguing informally here.

Since $\ln i^{1/i} =\frac{\ln i}{i} $, the sum is $\sum_{i=1}^n \frac{\ln i}{i}$.

Since $(\ln^2 x)' =2 \frac{\ln x}{x} $, $\int \frac{\ln x\ dx}{x} =\frac12 \ln^2 x $.

We then get $\sum_{i=1}^n \frac{\ln i}{i} \sim \int_1^n \frac{\ln x\ dx}{x} =\frac12 \ln^2 x \big|_1^n \sim \frac12 \ln^2 n $, so the limit is $\frac12$.

Answer 2

hint: $$\int \dfrac{1}{x} ln(x)={ln^2(x)\over 2}$$ and

$$ \int_2^nf(x)dx\geq\sum_2^n f(n)\geq \int_2^n f(x-1) dx$$ for $f(x)=\dfrac {ln^2(x)} 2$ and by applying the squeze theorem we are done.