Reciprocal of 19 from Sum of Powers of 2 Backwards

Theorem
The decimal expansion of the reciprocal of $19$ can be constructed by summing the powers of $2$, offset progressively backwards by $1$ digit:

1                       2                       4                      8                    16                   32                  64                128               256              512            1024           2048          4096         8192       16384      32768     65536   131072 + 262144 ...... -- .......1052631578947368421

while:
 * $\dfrac 1 {19} = 0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$

Proof
We confirm that from Reciprocal of $19$:
 * $\dfrac 1 {19} = 0 \cdotp \dot 05263 \, 15789 \, 47368 \, 42 \dot 1$

The construction above can be expressed as the sum:
 * $\ds \sum_{k \mathop \ge 0} \paren {2^k \times 10^k} = \sum_{k \mathop \ge 0} 20^k$

Obviously this sum does not converge.

However we can still analyse the last $N$ digits from this construction for every positive integer $N$.

From $k \ge N$ onwards, the last $N$ digits will not be affected by further additions of $20^k$ because:
 * $\forall k \ge N: 20^k \equiv 20^{k - N} \times 2^N \times 10^N \equiv 0 \pmod {10^N}$

By Number times Recurring Part of Reciprocal gives 9-Repdigit/Generalization, the last $N$ digits of this number coincides with the decimal expansion of the reciprocal of $19$ :
 * $\ds \sum_{k \mathop = 0}^{N - 1} 20^k \times 19 \equiv -1 \pmod {10^N}$

We have:

Hence the result.