The sum converges to 1000.
The [tex]n[/tex]-th partial sum of the series is
[tex]S_n=\displaystyle\sum_{i=1}^n100\left(\dfrac9{10}\right)^{i-1}=100\left(1+\dfrac9{10}+\left(\dfrac9{10}\right)^2+\cdots+\left(\dfrac9{10}\right)^{n-1}\right)[/tex]
Then
[tex]\dfrac9{10}S_n=100\left(\dfrac9{10}+\left(\dfrac9{10}\right)^2+\left(\dfrac9{10}\right)^3+\cdots+\left(\dfrac9{10}\right)^n\right)[/tex]
so that
[tex]S_n-\dfrac9{10}S_n=\dfrac1{10}S_n=100\left(1-\left(\dfrac9{10}\right)^n\right)[/tex]
[tex]\implies S_n=1000\left(1-\left(\dfrac9{10}\right)^n\right)[/tex]
As [tex]n\to\infty[/tex], [tex]\left(\dfrac9{10}\right)^n\to0[/tex], so we're left with
[tex]\displaystyle\sum_{i=1}^\infty100\left(\dfrac9{10}\right)^{i-1}=\lim_{n\to\infty}S_n=1000[/tex]
