Question

The sum of 30 times (1/3)^(n-1) from 1 to infinity

Answer 1

Let

`S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}`

Then

`\frac13S_n=\displaystyle\frac13+\frac1{3^2}+\frac1{3^3}+\cdots+\frac1{3^(n+1)}`

and

`S_n-\frac13S_n=\frac23S_n=1-\frac1{3^(n+1)}\implies S_n=\frac32-\frac1{2\cdot3^n}`

and as
`n\to\infty`, we end up with

`\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`

So we have

`\displaystyle\sum_(n=1)^\infty30\left(\frac13\right)^(n-1)=30\cdot\frac32=45`