Take any series that leads to a rational value.

$$\sum\limits_{k=0}^{\infty} a_{k} \in \mathbb{Q}, a_{k}\neq 0$$

Let us create a rational field extension over some transcendental number $z$, $\mathbb{Q}[z]=\{p+zq: p,q \in \mathbb{Q} \}$. We know that we can create an infinite number of *different* extensions of this type. For example, we can take $z_{k}=\pi^{\frac{1}{k}}, k \in \mathbb{N}, k > 0$.

Every extensions $\mathbb{Q}[z_{m}]$ contains values that are as close as we want to any $a_{n}$. We can associate one of $\mathbb{Q}[z_{m}]$ to each $a_{n}$. With that we have got a ground for possible substitution.

We take one $$b_{k} \in \mathbb{Q}[z_{k}]$$ and replace $a_{k}$ with that value.

With that we have replaced our series with

$$\sum\limits_{k=0}^{\infty} b_{k}$$

We need to prove that we can make $$\sum\limits_{k=0}^{\infty} b_{k} = \sum\limits_{k=0}^{\infty} a_{k}$$

However, $b_{k}$ can be as close as we want to $a_{k}$. With that we can make $\sum\limits_{k=0}^{\infty} a_{k}-b_{k}$ as small as we want, which means that for every extension $\mathbb{Q}[z_{k}]$ we can find $p_{k}$ and $q_{k} \neq 0$ so that $p_{k}+q_{k}z_{k}$ is a sufficiently good replacement for $a_{k}$ which will keep $\sum\limits_{k=0}^{\infty} b_{k}=\sum\limits_{k=0}^{\infty} a_{k}$

$p_{k}+q_{k}z_{k}$ from different extensions, where $q_{k} \neq 0$, are independent, because all values that are dependent with $q \neq 0$ are contained within each $\mathbb{Q}[z_{k}]$.

This means that $\sum\limits_{k=0}^{\infty} b_{k} \in \mathbb{Q}$ each $b_{k}$ is irrational (specifically transcendental) and any $b_{k}$ is independent as required.

(There would be no difference to make the extension over irrational numbers, transcendentals are making it all more obvious.)