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Is there any sequence $a_n$ of transcendental numbers, such that $ma_i\neq na_j$ for all integers m,n,i,j and any partial sum of $a_k$ is transcendental, but the total sum is rational ?

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    @anon: If it's relevant, it's relevant. No shame to be had in that! :)2011-10-30

3 Answers 3

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Pick any rational $r$ you'd like for the sum, then choose rationals $r_1,r_2,\dots$ so that $\sum a_n=r$,

where $a_n=r_n/\pi^n$.

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Yes. Let $\{\alpha_i\}$ be a transcendence basis for $\mathbb{R}$ over the algebraic closure of $\mathbb{Q}$, and we may assume they are all positive. Pick any $\alpha$, and then select $n\in\mathbb{N}$ such that $\frac{1}{2}\lt\alpha/n\lt 1$. Let $a_1= \alpha/n$.

Then pick $\beta\in\{\alpha_i\}-\{\alpha\}$, and find $n\in\mathbb{N}$ such that $\frac{1}{2}(1-a_1)\lt \frac{\beta}{n} \lt 1-a_1,$ and let $a_2 = \frac{\beta}{n}$.

Then pick $\gamma\notin\{\alpha_i\}-\{\alpha,\beta\}$, and find $n$ with $\frac{1}{2}(1-(a_1+a_2)) \lt \frac{\gamma}{n} \lt 1-(a_1+a_2)$ and let $a_3=\frac{\gamma}{n}$.

Lather, rinse, repeat.

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$a_n=\frac{1}{e}\frac{1}{n!}$ works.

Any finite sum of such terms is of the type $\frac{1}{e}$ times a rational number, while the total sum $\sum_{n=0}^\infty a_n =1$.

Revised

$a_n=\frac{1}{e^n} (1-\frac{1}{e})=\frac{e-1}{e^{n+1}}$. Again the sum is 1 and it is easy to argue that it satisfies all the requirements this time.

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    Missed that, will fix it in a moment :)2011-10-30