# Can we find a positive real sequence $(a_n)$ with $\sum_{n=1}^\infty a_n^k=1/k$ for all positive integer $k$?

Suppose that $$a_1, a_2, a_3, \ldots$$ are positive real numbers such that \begin{align} 1/k=\sum_{n=1}^\infty a_n^k \qquad \text{ for all integers } k>1. \end{align} What are $$a_1, a_2, a_3, \ldots$$?

We must have $$0 for all $$n$$. Also note that for $$k=2$$, we must have for all $$n$$ the inequality $$\displaystyle a_n^2\leq \frac{1}{2}$$ hence $$\displaystyle a_n\leq \frac{1}{\sqrt{2}}$$. We have then for large $$k$$: $$1-a_n\geq 1-\frac{1}{\sqrt{2}}>\left(\frac{1}{\sqrt{2}}\right)^{k^2}\geq a_n^{k^2}$$ and thus $$1-a_n-a_n^{k^2}> 0$$ for all $$n$$ and large $$k$$.
Now note that $$\displaystyle \frac{1}{k}=\frac{1}{k+1}+\frac{1}{k(k+1)}$$. Hence
$$\sum_{n\geq 1} a_n^k=\sum_{n\geq 1} a_n^{k+1}+\sum_{n\geq 1} a_n^{k(k+1)}$$ We have hence for all (large) $$k\geq 1$$ $$\sum_{n\geq 1}a_n^k(1-a_n-a_n^{k^2})=0$$ and as $$a_n^k(1-a_n-a_n^{k^2})>0$$ for all $$n$$, we have a contradiction, no such sequence can exists.