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 $n$?

Question

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.

Answer 1

Let $M$ be the product of the first $N$ primes and for $0\le r<M$ define $$b_r=\sum_{k\equiv r\pmod M}a_k\ge0$$ (without any convergence problems). Then $$ b_0+b_d+b_{2d}+\cdots+b_{M-d}=\frac 1d$$ for all divisors $d$ of $M$. If we multiply each such equation by $\mu(d)$ and add, then we obtain $$ \sum_{k=0}^{M-1}c_kb_k=\sum_{d\mid M}\frac{\mu(d)}d=\prod_{i=1}^N\left(1-\frac1p\right)$$ where $$c_k=\sum_{d|k}\mu(d)=\begin{cases}1&k=1\\0&k>1\end{cases}$$ In other words, $$b_0= \prod_{i=1}^N\left(1-\frac1p\right).$$ On the other hand, $b_0=\frac 1M$ - a contradiction contradiction already for $M=2\cdot3\cdot 5$.

Answer 2

Let's put $$ f(z) = \sum\limits_{0\, \le \,j} {\,a_{\,j} z^{\,j} } $$

Then the application of the Series Multisection tells us that $$ \sum\limits_{0\, \le \,j} {\,a_{\,n\,j} z^{\,n\,j} } = {1 \over n}\sum\limits_{0\, \le \,k\, \le \,n - 1} {f(\omega ^{\,k} z)} $$

Therefore $$ \sum\limits_{0\, \le \,j} {\,a_{\,n\,j} z^{\,n\,j} } = {1 \over n}\sum\limits_{0\, \le \,k\, \le \,n - 1} {f(\omega _{\,n} ^{\,k} z)} \quad \left| {\;\omega _{\,n} = e^{\,i2\pi /n} } \right. $$ and we are looking for a function $f(z)$ such that $$ \sum\limits_{0\, \le \,k\, \le \,n - 1} {f(\omega _{\,n} ^{\,k} )} = 1 $$ which means that the function takes on values over the roots of $1^n$ which sum to $1$ for all values of $n$.

Now, the roots corresponding to $k/n = 1/2, 2/4, 3/6, ...$ have the same value, as all those for which $k/n = const.$.
Consequently also the function will have the same value on those roots.

I am not an expert in the theory of functions, but definitely $f(x)$ cannot be continuous.

Answer 3

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.

Answer 4

Many beautiful solutions have been presented here. But I think this would be the most elementary one.

Note: This solution is not mine.

We will prove that $a_i=0$ for all $i\ge 0$. $a_0=0$ has already been proved in previous answers. Begin by observing that $$\sum_{\gcd(i,n)=1} a_i=\frac{\phi(n)}{n}$$ if $n$ is a square-free integer. Now choose a particular $m$ and set $n$ to be relatively prime to $m$(For example, choose $m=8$ then set $n=5$). Thus the LHS is at-least $a_m$. But the RHS can be made arbitrarily close to 0 because $\prod_{i=1}^{\infty} p_i=0$ where $p_i$ stands for the $i$-th prime.