There seems to be a lot of confusion in the other answers and comments thereto. I'll try to make things more clear. (In particular, @Holowitz's and @Nico's answers are incorrect, as I showed in the comments below @Nico's answer.)

First note that a real number can be identified with a set of natural numbers by declaring that the binary expansion of the real corresponds to the characteristic sequence set of natural numbers. (Of course, some reals have non-unique binary expansion, so this identification doesn't always make sense, but any such real gives rise to a computable set no matter which binary expansion is taken, so insofar as we're concerned, this is irrelevant.)

Now, the definition of computable you gave is what's known as *limit computable* or $\Delta_{2}^{0}$, though since the notation $\Delta_{2}^{0}$ is *a priori* reserved for sets defined by formulas that are both $\Sigma_{2}^{0}$ and $\Pi^{0}_{2}$ in the arithmetical hierarchy, it must be proven that the two notions coincide. This is exactly Schoenfield's limit
lemma together with Post's theorem.

Regarding your request for a "concrete" example of a real that is not $\Delta^{0}_{2}$ (i.e. one that can be "picked by defining it"), the number--let's call it $\alpha$--given in the paper cited in the answer you accepted is defined only relative to some fixed (incomputable!) enumeration of all $\Delta^{0}_{2}$ reals; different enumerations give rise to different $\alpha$'s. However, such an enumeration is inherently not $\Delta^{0}_{2}$ (it can be $\Delta^{0}_{3}$ as @Carl points out in the comments to @Mahmud's answer) in the arithmetical hierarchy, so I do think that it is an acceptable example (contrary to my comments) . Given that, though, it's evident that a much better (i.e. less complex) real can be given: take any strictly $\Sigma^{0}_{2}$ or $\Pi^{0}_{2}$ real. For example, the set of indices of total computable functions $\mathrm{Tot}:=\{e:\forall n \exists s \, [\varphi_{e,s}(n)\downarrow] \}$ is a well-known $\Pi^{0}_{2}$-complete real.

The moral of the story is that there are many arithmetically definable reals that are not "computable" in the sense you defined; just take any arithmetical real that is not $\Delta^{0}_{2}$.

No. For any $n$, define

$a_n =
\begin{cases}
\pi/2^r, & \text{if Turing machine $n$ halts in $r$ steps on input $n$} \\
0, & \text{if Turing machine $n$ never halts on input $n$}
\end{cases}
$

$a_n$ is computable; we can approximate it to any desired accuracy. But if you could determine whether $a_n$ was transcendental or not, you would have solved the Halting Problem.

## Best Answer

Using this definition of left-computability, our aim is to construct a computable increasing sequence of rationals that converges to a non-computable number.

Consider a sequence $q_{n}$ subunitary rational numbers.

Define the $k$'th digit after the decimal point of $q_{n}$ to be $1$ if $k \leq n$ and the program encoded by $k$ halts after $n$ steps or less, or $0$ otherwise.

The $q_{n}$ are increasing, as a digit that becomes $1$ stays $1$ for all $q_{p}$ with $p>n$. They are also rational, as each $q_{n}$ has at most $n$ non-zero decimal places. However, their limit will be a number $q$ who's $k$th digit is $1$ if the program encoded by $k$ halts, or $0$ otherwise. As it encodes the solution to the halting problem, it is not computable.

Thus $q$ is left-computable, but not computable.