[tex]7+0.7+0.007+\cdots=7\left(1+\dfrac1{10}+\dfrac1{100}+\cdots\right)[/tex]
Let [tex]S_n[/tex] denote the [tex]n[/tex]th partial sum of the series, i.e.
[tex]S_n=1+\dfrac1{10}+\dfrac1{100}+\cdots+\dfrac1{100^n}[/tex]
Then
[tex]\dfrac1{10}S_n=\dfrac1{10}+\dfrac1{100}+\dfrac1{1000}+\dfrac1{10^{n+1}}[/tex]
and subtracting from [tex]S_n[/tex] we get
[tex]S_n-\dfrac1{10}S_n=\dfrac9{10}S_n=1-\dfrac1{10^{n+1}}[/tex]
[tex]\implies S_n=\dfrac{10}9\left(1-\dfrac1{10^{n+1}}\right)[/tex]
As [tex]n\to\infty[/tex], the exponential term vanishes, leaving us with
[tex]\displaystyle\lim_{n\to\infty}S_n=\dfrac{10}9[/tex]
and so
[tex]7+0.7+0.007+\cdots=7.777\ldots=\dfrac{70}9[/tex]
