I want to evaluate the following limit
\begin{equation} \lim \limits_{n \to \infty} \sum \limits_{i=0}^{n-1} \frac{\sqrt{i \ln(n-i)}}{n^{3/2}} \end{equation}
I was thinking about rewriting it as Riemann sum and estimating the limit trough integrals. Does this limit exist? How would I rewrite it as a Riemann sum? Thank you for your help.
Rewrite the sum as
$$ \sum_{i = 0}^{n-1} \frac{1}{n} \sqrt{\frac{i}{n}} \sqrt{\ln(n-i)} \geq \sum_{i = 0}^{\lfloor n/2\rfloor} \frac{1}{n} \sqrt{\frac{i}{n}} \sqrt{\ln(n-i)} \geq \sqrt{\ln(n/2)} \sum_{i = 0}^{\lfloor n/2\rfloor} \frac{1}{n} \sqrt{\frac{i}{n}} $$
The sum on the far right converges to $\int_{0}^{1/2} \sqrt{x} ~\mathrm{d}x $ which is a fixed constant. But the factor $\sqrt{\ln(n/2)}$ grows unboundedly, and hence your original sequence diverges.
If you don't want to use the Riemann sum approximation, first notice that the summands are positive. So it is bounded below by the sum from $n/4$ to $n/2$. This means that your original sum is bounded below by
$$ \sum_{i = n/4}^{n/2} \frac{\sqrt{n/4 \cdot \ln(n/2)}}{n^{3/2}} \geq \frac{1}{8} \sqrt{\ln(n/2)} $$
and so diverges as $n\to \infty$.