In [Show that $\sum\limits\_pa\_p$ converges iff $\sum\limits\_{n}\frac{a\_n}{\log n}$ converges](https://math.stackexchange.com/questions/1458492/show-that-sum-limits-pa-p-converges-iff-sum-limits-n-fraca-n-log-n-c) it is said that this sequence of partial sums converges
$$
\begin{split}
\sum\_{1<n\leq N}\frac{a\_{n}}{\log\left(n\right)} &=\sum\_{1<n\leq N}1\cdot\frac{a\_{n}}{\log\left(n\right)} \\
& =\frac{\left(N-1\right)a\_{N}}{\log\left(N\right)}+\sum\_{k\leq N-1}\left(k-1\right)\left(\frac{a\_{k}}{\log\left(k\right)}-\frac{a\_{k+1}}{\log\left(k+1\right)}\right)
\end{split}
$$ iff this sequence of partial sums converges
$$\sum\_{p\leq N}a\_{p}=\pi\left(N\right)a\_{N}+\sum\_{k\leq N-1}\pi\left(k\right)\left(a\_{k}-a\_{k+1}\right)∼ \frac{Na\_{N}}{\log\left(N\right)}+\sum\_{k\leq N-1}\frac{k}{\log\left(k\right)}\left(a\_{k}-a\_{k+1}\right)$$ But I don't see why. I think this should be clear but I can't get it. Can someone explain? ($a\_n$ is a non-increasing sequence of positive numbers.)



One series converges iff the other converges
Since $a\_n$ is nonincreasing and nonnegative, $a\_n$ converges to some real $a\ge0$. If $a>0$, then neither one of the two series converges. It remains to consider the case $a=0$. Then
\begin{equation\*}
 a\_n=\sum\_{j\ge n}b\_j \tag{1}\label{1}
\end{equation\*}
for some nonnegative $b\_j$'s.


By the prime number theorem,
\begin{equation\*}
 \sum\_{k\le n} 1(k\in P)\sim \frac n{\ln n}\sim\sum\_{k\le n}\frac1{\ln k}, \tag{2}\label{2}
\end{equation\*}
where $P$ is the set of all prime numbers.


In view of \eqref{1},
\begin{equation\*}
\begin{aligned}
 \sum\_p a\_p&=\sum\_n 1(n\in P)\,a\_n \\ 
 &=\sum\_n 1(n\in P)\,\sum\_{j\ge n}b\_j \\ 
 &=\sum\_j b\_j\sum\_{n\le j} 1(n\in P). 
\end{aligned} 
\tag{3}\label{3}
\end{equation\*}


By \eqref{2},
\begin{equation}
 \sum\_j b\_j\sum\_{n\le j} 1(n\in P)<\infty \iff 
 \sum\_j b\_j\sum\_{k\le j}\frac1{\ln k}<\infty. 
 \tag{4}\label{4}
\end{equation}
But
\begin{equation\*}
\begin{aligned}
 &\sum\_j b\_j\sum\_{k\le j}\frac1{\ln k} \\ 
 &=\sum\_k\frac1{\ln k}\sum\_{j\ge k} b\_j \\ 
 &=\sum\_k\frac{a\_k}{\ln k}. 
 \end{aligned}
 \tag{5}\label{5} 
\end{equation\*}


It follows from \eqref{3}, \eqref{4}, and \eqref{5} that  

\begin{equation}
 \sum\_p a\_p<\infty \iff 
\sum\_k\frac{a\_k}{\ln k}<\infty, 
\end{equation}
as desired.


