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 $0<a_n<1$ 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.

  • 2
    $\begingroup$ awesome! Very nice solution! $\endgroup$ – Mohammad Khosravi Jul 21 '14 at 20:13

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