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Notice that the given sum is equivalent to
[tex]10 + 1 + \dfrac1{10} + \dfrac1{10^2} + \dfrac1{10^3} + \cdots \\\\ ~~~~~~~~~~~~ = 10 \left(1 + \dfrac1{10} + \dfrac1{10^2} + \dfrac1{10^3} + \dfrac1{10^4} + \cdots\right)[/tex]
or 10 times the infinite geometric series with ratio 1/10.
To compute the infinite sum, consider the [tex]n[/tex]-th partial sum
[tex]S_n = 1 + \dfrac1{10} + \dfrac1{10^2} + \cdots + \dfrac1{10^{n-1}}[/tex]
Multiply both sides by the ratio.
[tex]\dfrac1{10} S_n = \dfrac1{10} + \dfrac1{10^2} + \dfrac1{10^3} + \cdots + \dfrac1{10^n}[/tex]
Subtract this from [tex]S_n[/tex] to eliminate all but the outermost terms,
[tex]S_n - \dfrac1{10} S_n = 1 - \dfrac1{10^n} \implies S_n = \dfrac{10}9 \left(1 - \dfrac1{10^n}\right)[/tex]
As [tex]n\to\infty[/tex], the exponential term will decay to 0, leaving us with
[tex]1 + \dfrac1{10} + \dfrac1{10^2} + \cdots = \dfrac{10}9[/tex]
Then the value of the sum we want is
[tex]10\times\dfrac{10}9 = \boxed{\dfrac{100}9}[/tex]