Can an infinite sum of non-computable numbers be computable, such that all finite sums of subsets of terms are non-computable?

Question

Background

In the following question, User1 asks whether an infinite sum of irrational numbers can be rational. Multiple answers¹ indicate the answer to this question is 'yes'. For instance, Rasmus Erlemann notes that

$$\tan \frac{\pi}4=\sum_{n=0}^\infty \frac{(-1)^n 2^{2n+2}(2^{2n+2}-1)B_{2n+2}}{(2n+2)!}\left(\frac{\pi}4\right)^{2n+1}=1, $$ by the Maclaurin series of $\tan(\cdot)$.

The Question

This question is asked in a similar spirit. I wonder whether there are any infinite series consisting solely of non-computable terms, that amount to a computable number. Examples of such non-computable numbers can found over here. They include Chaitin's constants.

To make the question more precise, define $$ C = \sum_{k=1}^{\infty} u_{k} . $$ Here, all $u_{k}$ are non-computable numbers. I am looking for explicit examples of infinite series such that $C$ is a computable number, and all $u_{k}$ are both positive and linearly independent over $\mathbb{Q}$.

Added

Although user TonyK answers my initial question correctly, I would like to add another restriction to make the question more interesting. I furthermore require that every sum of a finite subset of summands of the infinite series is also non-computable. This makes TonyK's example unworkable, since, for instance, the sum of the first two terms is computable: $$ \alpha u_{1} + (1-\alpha)u_{1} = u_{1}, $$ where $\alpha$ is non-computable and $u_{1}$ is a positive computable term.

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^Notes

^{[1] If you're interested in such series, I encourage you to also take a look at my recent answer. It involves a class of representations of numbers, called rational zeta series, that offers a fruitful way to approach this problem. For instance, the answer includes the series identity $$\sum_{n=1}^{\infty} \left[ \zeta(2n)-1 \right] = \frac{3}{4} . $$}

Answer 1

Let $\alpha\in(0,1)$ be an uncomputable number, and let $\sum u_k$ be any convergent series of positive computable terms linearly independent over $\Bbb Q$. Then the series $$\alpha u_1+(1-\alpha)u_1+\alpha u_2+(1-\alpha)u_2+\cdots$$ satisfies your conditions (because any linear dependency over $\Bbb Q$ would give us a way to compute the uncomputable $\alpha$).

You asked for an explicit example, so take for instance $u_k=p_k^{-3/2}$, where $p^k$ is the $p$th prime.