Transcendentals that sum to rationals 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 ?
 A: 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 $a_n=r_n/\pi^n$.  
A: 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. 
A: $a_n=\frac{1}{e}\frac{1}{n!}$ 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.
