Let
[tex]S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}[/tex]
Then
[tex]\dfrac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^{n+1}}[/tex]
and
[tex]S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}[/tex]
and as [tex]n\to\infty[/tex], we end up with
[tex]\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32[/tex]
So we have
[tex]\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45[/tex]
