Question

1/5+1/15+1/45+1/81+... Converges or diverges. Sum or no sum

Answer 1

Let's consider the
`n` partial sum of the series:


`S_n=\frac15+\frac1{15}+\frac1{45}+\frac1{81}+\cdots+\frac1{5*3^(n-1)}`

`S_n=\frac15\left(1+\frac13+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}+\cdots+\frac1{3^(n-1)}\right)`

Multiply both sides by
`\frac13`, making sure to distribute
`\frac13` to each term in the sum on the right side:


`\frac13S_n=\frac15\left(\frac13+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}+\frac1{3^5}+\cdots+\frac1{3^n}\right)`

Now subtract this from
`S_n` to get


`S_n-\frac13S_n=\frac23S_n=\frac15\left(1-\frac1{3^n}\right)`

`\implies S_n=\frac3{10}\left(1-\frac1{3^n}\right)=\frac3{10}-\frac1{10*3^(n-1)}`

Now as
`n\to\infty`, the second term converges to 0, leaving you with


`\displaystyle\lim_(n\to\infty)S_n=\frac3{10}`