<s>[INST] Are these proofs logically equivalent? Here are two proofs, firstly:
x = 0.999...
10x = 9.999...
    = 9 + 0.999...
    = 9 + x
9x = 9
x = 1

And secondly:
x = 1 - 1 + 1 - 1 + 1 - 1 ...
  = 1 - (1 - 1 + 1 - 1 + 1 ...
  = 1 - x
2x = 1
x = 1/2

The fourth line in the first and the third line in the second use the same trick. Does the legitimacy of the first proof guarantee that of the second?
 [/INST] No.  The first series is absolutely convergent, so its terms can be rearranged without changing the sum, while the second series is divergent.  
The number $0.\overline{9}$ can be expressed in the form 
\begin{align*}
0.\overline{9} & = \sum_{k = 1}^{\infty} \frac{9}{10^k}\\
               & = \frac{9}{10} \sum_{k = 0}^{\infty} \frac{1}{10^k}
\end{align*}
which is a convergent geometric series, so 
\begin{align*}
0.\overline{9} & = \frac{9}{10} \cdot \frac{1}{1 - \frac{1}{10}}\\
               & = \frac{9}{10} \cdot \frac{1}{\dfrac{9}{10}}\\
               & = \frac{9}{10} \cdot \frac{10}{9}\\
               & = 1
\end{align*}
In fact, the series 
$$0.\overline{9} = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$
is absolutely convergent since the series 
$$\sum_{k = 1}^{\infty} \bigg| \frac{9}{10^K} \bigg| = \sum_{k = 1}^{\infty} \frac{9}{10^k}$$
converges.  Any rearrangement of an absolutely convergent series has the same sum.
The series 
$$1 - 1 + 1 - 1 + 1 - 1 + \cdots = \sum_{k = 0}^{\infty} (-1)^k$$
diverges since the sequence of partial sums $\{1, 0, 1, 0, 1, 0, \ldots\}$
alternates between $1$ and $0$.  Since the series never converges to a limit, it is meaningless to say that 
$$x = 1 - 1 + 1 - 1 + 1 - 1 + \cdots$$
A series must be absolutely convergent to guarantee that rearranging the terms of the series does not change the sum.
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