discrete not continuous

Firstly we see that the series $\displaystyle \sum_{i=1}^{\infty} \frac{\ln(i)}{i}$ is divergent(By Cauchy Condensation Test ) and hence the sequence of partial sums which is $\displaystyle S_{n}=\sum_{i=1}^{n} \frac{\ln(i)}{i}$ must be a divergent sequence and it diverges to positive infinity.

Also the sequence $\displaystyle (\ln(n))^{2}$ is monotone , divergent and tends to positive infinity.

So we can apply the Stolz-Cesaro Theorem :-

So we have to solve the following limit:- $\displaystyle \lim_{n\to\infty} \frac{S_{n+1}-S_{n}}{(\ln(n+1))^{2}-(\ln(n))^{2}} =\lim_{n\to\infty} \frac{\frac{\ln(n+1)}{n+1}}{\left(\ln(n+1)-\ln(n)\right)\left(\ln(n+1)+\ln(n)\right)}$

Notice that this is neither in $\displaystyle \frac{0}{0}$ form nor the $\displaystyle \frac{\infty}{\infty}$ form . Hence you cannot apply L'Hopital or Stolz Cesaro again without making certain rearrangements first. Rearranging the expression a little we get:-

$\displaystyle \lim_{n\to\infty}\frac{1}{(n+1)\ln(1+\frac{1}{n})}\cdot \frac{\ln(n+1)}{\ln(n)+\ln(n+1)}$

Now both of the sequences :- $\displaystyle \frac{1}{(n+1)\ln(1+\frac{1}{n})}$ and $\displaystyle \frac{\ln(n+1)}{\ln(n)+\ln(n+1)}$ are convergent and so we can evaluate the limits separately and then multiply them to get the answer.

$\displaystyle \lim_{n\to\infty} \frac{1}{(n+1)\ln(1+\frac{1}{n})} = \lim_{n\to\infty} \frac{1}{(n+1)\left(\frac{1}{n} - \frac{1}{2n^{2}} -O(\frac{1}{n^{3}})\right)} = 1$

Again for the sequence $\displaystyle \lim_{n\to\infty}\frac{\ln(n+1)}{\ln(n)+\ln(n+1)}= \frac{1}{2}$ . For the proof you can apply L'Hopital rule . But I don't think you need to. it is enough to just consider $\displaystyle \lim_{n\to\infty} \frac{\ln(n+1)}{\ln(n)} = 1$ . This is achieved immediately by applying Sandwich Theorem .

Now the answer follows directly and it is $\displaystyle \frac{1}{2}$

By the integral test inequality by have : $\sum_{i=1}^n \frac{\log i}{i} \sim \int_1^n \frac{\log x}{x} \ \mathrm{d}x = \frac{\log^2 n}{2}$ The limit comes directly.