<s>[INST] Can an infinite sum of irrational numbers be rational? Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational.
Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear combination of the other ?
By linear combination, we mean there exists some rational numbers $u,v$ such that $a_i = ua_j + v$.
 [/INST] Consider that $$\frac{1}{1-x} = 1 + x + x^{2} + x^{3} + \dots  \text{ (for } -1 < x < 1).$$
Then since $\frac{\pi}{5}$ is in $(-1,1)$, we have:
$$\underbrace{\frac{1}{1 - \frac{\pi}{5}}}_{\dfrac{5}{5-\pi}} = 1 + \underbrace{\frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots}_{\text{infinite sum of irrational numbers}}$$
Subtracting $1$ from both sides of the above equation gives:
$$\underbrace{\dfrac{5}{5-\pi} - 1}_{\dfrac{\pi}{5-\pi}} =  \frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots$$
Now the right hand side is an infinite sum of terms with each term an irrational number.  Unfortunately, the left hand side is also irrational. Let's multiply both sides of the above equation by the reciprocal of the left hand side to induce rationality of that side, i.e., multiply both sides by $\frac{5-\pi}{\pi}$.  This gives:
$$\frac{5-\pi}{\pi}\cdot \dfrac{\pi}{5-\pi}=  \frac{5-\pi}{\pi} \cdot \left [\frac{\pi}{5} + \frac{\pi^{2}}{5^{2}} + \frac{\pi^{3}}{5^{3}} + \dots \right ]$$
Simplifying gives:
$$\underbrace{1}_{\text{rational number}} = \underbrace{\frac{5-\pi}{5} + \frac{(5-\pi)\pi}{5^{2}} + \frac{(5-\pi)\pi^{2}}{5^{3}} + \dots}_{\text{infinite sum of irrational terms}} $$
EDIT: Further justification/verification that the sum in the right hand side of the equation above equals $1$: Recognize it as a geometric series!  The factor we multiply at each step is $r:=\frac{\pi}{5}$.  Then we know the series converges to $\frac{\text{first term}}{1 - r} = \dfrac{\frac{5-\pi}{5}}{1 - \frac{\pi}{5}} = \dfrac{\frac{5-\pi}{5}}{\frac{5-\pi}{5}} = 1$.
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