Does there exist a sequence $\{a_n\}_{n \ge 0}$ of nonnegative reals such that $ \sum_{j \ge 0} a_{nj} = \frac{1}{n}$ holds for all naturals $n$? Does there exist a sequence $\{a_n\}_{n \ge 0}$ of nonnegative reals such that
$$ \sum_{j \ge 0} a_{nj} = \dfrac{1}{n}$$
holds for all naturals $n$?
My progress: I could show that $a_n\le \frac{1}{2n}$. I am not sure if this is even useful.

Let $$a_{n,j}=\frac{1}{n+j}-\frac{1}{n+j+1}\geqslant 0$$
It is a telescopic sum :
$$ \sum_{j=0}^{+\infty}a_{n,j}=\sum_{j=0}^{+\infty}\left(\frac{1}{n+j}-\frac{1}{n+j+1}\right)=\frac{1}{n} $$
EDIT : If you have $\sum_{j=0}^{+\infty}a_{nj}=\frac{1}{n}$ for all $n\in\mathbb{N}^*$. First notice that $a_0=0$ because for $n$ large enough, you would have $\frac{1}{n}<a_0$ and thus $a_0\leqslant\sum_{j=0}^{+\infty}a_{nj}=\frac{1}{n}<a_0$. Let $\mu$ be the Möbius function, first
$$ \sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=\sum_{(n,j)\in\mathbb{N}^*\times\mathbb{N}^*} \mu(n)a_{nj}=\sum_{p=1}^{+\infty}a_p\sum_{d|p}\mu(d)=a_1$$
because $\sum_{d|p}\mu(d)=0$ if $p\geqslant 2$ and $1$ if $p=1$. But because of the prime numbers theorem we have
$$ \sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=0 $$
(Go here for a proof : https://books.google.fr/books?id=Il64dZELHEIC&pg=PA94&redir_esc=y#v=onepage&q&f=false )
thus $a_1=0$. Now let $\left(b_n^{(p)}\right)=(a_{np})$ for all $p\in\mathbb{N}^*$, we have
$$ \forall n\in\mathbb{N}^*,\sum_{j=1}^{+\infty}b_{nj}^{(p)}=\sum_{j=1}^{+\infty}a_{npj}=\frac{1}{np} $$
Thus $a_p=b_1^{(p)}=\sum_{n=1}^{+\infty}\frac{\mu(n)}{np}=\frac{a_1}{p}=0$ for all $p\in\mathbb{N}^*$ which is not. Thus such a sequence can't exist.
