Convergence of $\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$

On one hand since $\frac {1}{n}\le \frac 1n\left(1+\frac 12+\ldots+\frac 1n\right)$ it is not difficult to see that for $p>0$ we have

$$\sum_{n=0}^{\infty} \frac {1}{n^p}\le\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$ so the divergence statement from Riemann series which diverges for $p\le 1$

On the other hand it follows from Hardy inequality that: for $p>1$ $$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p\le \left(\frac{p}{p-1}\right)^p\sum_{n=0}^{\infty} \frac {1}{n^p}<\infty$$ the converges blatantly follows from the Riemann series on the right hand side which converges only for $p>1.$

By Riemann sums, we have $\log n\lt\sum\limits_{i=1}^n\frac1i\lt1+\log n$
This leads to $\sum \frac1n(\log n)^p$, which, by another Riemann sum, can be compared with $\int_1^n\frac1x(\log x)^pdx=\int_0^{\log x}y^pdp$