We can employ a simple repeated decimal trick:
[tex]x=1.111\ldots[/tex]
[tex]0.1x=0.111\ldots[/tex]
[tex]\implies x-0.1x=1\implies0.9x=1\implies x=\dfrac1{0.9}=\dfrac{10}9[/tex]
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Alternatively, we can compute the partial sum of the series.
[tex]\displaystyle S_n=\sum_{k=0}^n\dfrac1{10^k}[/tex]
[tex]S_n=1+0.1+0.01+\cdots+\dfrac1{10^n}[/tex]
[tex]0.1S_n=0.1+0.01+0.001+\cdots+\dfrac1{10^{n+1}}[/tex]
[tex]\implies S_n-0.1S_n=0.9S_n=1-\dfrac1{10^{n+1}}[/tex]
[tex]\implies S_n=\dfrac{10}9-\dfrac9{10^n}[/tex]
As [tex]n\to\infty[/tex], the second term vanishes and we're left with [tex]\dfrac{10}9[/tex]. Notice that this is really just a more formal version of the earlier trick.
