4
$\begingroup$

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 $a_1, a_2, a_3, \ldots$?

$\endgroup$
13
$\begingroup$

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.