Prove that $$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$ converges if $p>1$ and diverges when $0<p\leqslant1$.
My attempt:
Since the terms of the series $\sum_{n=0}^{\infty} \frac1n$ are monotonically decreasing and positive, I applied Cauchy's Condensation theorem, i.e, if $\sum a(n)$ and $\sum2^na(2^n)$ will converge and diverge together. Applying this I got, $\sum\left(1-\frac{1}{2^n}\right)^p$, but I am stuck now. Don't know how to proceed further.
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 $\int_1^n\frac1x(\log x)^pdx=\int_0^{\log x}y^pdp$
In $$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$
$$1+\frac 12+\ldots+\frac 1n=O(\log(n))$$
so , given sum equals
$$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p=\sum_{n=0}^{\infty}\left(\frac{\log(n)}{n}\right)^p$$
Here, $O(n)\gg O(\log(n))$ ,
so convergence of given sum depends on convergence of $$\sum_{n=0}^{\infty}\left(\frac{1}{n}\right)^p$$
which converges if $p>1$ and diverges when $0<p\leq1$.
For case $p=1$
given sum equals $$\sum_{n=0}^{\infty}\left(\frac{\log(n)}{n}\right)$$ which is divergent as
$$\sum_{n=0}^{\infty}\left(\frac{1}{n(\log^p(n))}\right)$$
converges if $p>1$
