Comparing with $\sum\frac{1}{n}$ to distinguish the convergence and divergence of series of positive terms Considering the series
$$
\sum_{n=1}^\infty \frac{1}{n^p},
$$
since when $p> 1$ the series is  convergent, and when $0<p\le 1$ it is divergent, I want to construct comparison theorem  to distinguish the convergence and divergence  of series. For example:
Assuming $\sum a_n$ is a series of positive terms, then $\sum a_n$ is convergent iff $\exists \eta >0, C>0$ such that
$$
|a_{\eta+\delta }|< \frac{C}{\delta},~~~\forall \delta \in \mathbb Z^+.
$$
Obviously, it is wrong.  I want to distinguish the convergence and divergence of series by comparing with $\sum \frac{1}{n}$. But I fail to find a suitable result.
 A: You can use the Cauchy Condensation test on $a_n = \frac{1}{n}$. Note that $\{a_n\}_{n = 1}^\infty$ is a decreasing sequence.

$\bullet$ Cauchy Condensation Test: Consider the following chain
\begin{align*}
    a_{1} \geqslant a_{2} \geqslant \cdots \geqslant 0
\end{align*}
Or in other words, the non-negative real sequence $\{a_{n}\}_{n = 1}^{\infty}$ is decreasing. Then the series
\begin{align*}
    \sum_{n = 1}^{\infty} a_{n} = a_{1} + a_{2} + \cdots
\end{align*}
respectively converges or diverges if and only if the series
\begin{align*}
    \sum_{k = 0}^{\infty} 2^{k}a_{2^{k}} = a_{0} + 2 a_{2} + 4 a_{4} + \cdots 
\end{align*}
converges or diverges.

Then we have that
$$ \sum_{k = 0}^\infty \frac{2^k}{2^{kp}} = \sum_{k = 1}^\infty \frac{1}{2^{kp - k}}  $$
Then by root test
$$ \sqrt[n]{\frac{1}{2^{np - n}}} = \frac{1}{2^{p-1}} $$
Now we can conclude that
$$ \sqrt[n]{2^na_{2^n}} = \frac{1}{2^{p-1}} = \begin{cases}
< 1 & \text{if } p > 1\\
= 1 & \text{if } p = 1\\
> 1 & \text{if } p < 1
\end{cases} $$
Thus convergent when $p > 1$, divergent when $p \leqslant 1$

$\color{purple}{\textbf{Note:}}$ Note that from the ratio test we can't conclude anything for $p=1$ case. But we know that $$ \sum_{n = 1}^\infty \frac{1}{n} = \infty $$
So, this is just a note.
