Convergence of $\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$ Prove that $$\sum_{n=0}^{\infty}  \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p$$ converges if $p&gt;1$ and diverges when $0&lt;p\leqslant1$.


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 $p&gt;0$ 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 $p&gt;1$
$$\sum_{n=0}^{\infty} \left(\frac{1+\frac 12+\ldots+\frac 1n}{n}\right)^p\le \left(\frac{p}{p-1}\right)^p\sum_{n=0}^{\infty} \frac {1}{n^p}&lt;\infty$$
 the converges blatantly follows from the Riemann series on the right hand side which converges only for $p&gt;1.$
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, $O(n)&gt;&gt;O(log(n))$ , 

so convergence of given sum depends on convergence of $$\sum_{n=0}^{\infty}\left(\frac{1}{n}\right)^p$$

which converges if $p&gt;1$ and diverges when $0&lt;p\leq1$.

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 $p&gt;1$