The series shows:

`1/3^1+ 1/3^2 + 1/3^3+...` . OR using the rules of exponents

`3^-1 + 3^-2 + 3^-3 ... ` .

`therefore T_(n) = 3^-n` or `T_(n) = (1/3)^n`

Use the sum for a geometric series as there is a constant ratio. To check the ratio

`T_(3)/ T_2 =...

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The series shows:

`1/3^1+ 1/3^2 + 1/3^3+...` . OR using the rules of exponents

`3^-1 + 3^-2 + 3^-3 ... ` .

`therefore T_(n) = 3^-n` or `T_(n) = (1/3)^n`

Use the sum for a geometric series as there is a constant ratio. To check the ratio

`T_(3)/ T_2 = 1/27 divide 1/9 = 1/3` which is the same as `T_(2)/T_(1) = 1/9 divide 1/3 = 1/3`

The first term is `a`

So `a=1/3 and r=1/3`

The formula is :`S_(n) = (a(1-r^n))/(1-r)`

Therefore the sum of the series is `S_(n) = (1/3(1-(1/3)^n))/(1-1/3)`

To write in sigma notation where n=n ( as we do not know how many terms) or** summation notation:**

`sum_(n=1)^n` `3^-n` `= (a(1-r^n))/(1-r)`