# `sum_(n=0)^oo (2/3)^n ` Determine the convergence or divergence of the series.

## Expert Answers

The given series `sum_(n=0)^oo (2/3)^n` is in a form of the geometric series.

Recall that the sum of geometric series follows the formula: `sum_(n=1)^oo a*r^(n-1)` .

or with an index shift: `sum_(n=0)^oo a*r^n = a+a*r + a*r^2 +...`

The convergence test for the geometric series follows the conditions:

a) If `|r|lt1`  or `-1 ltrlt 1` then the geometric series converges to `sum_(n=0)^oo a*r^n =sum_(n=1)^oo a*r^(n-1)= a/(1-r)` .

b) If `|r|gt=1` then the geometric series diverges.

By comparing   `sum_(n=0)^o(2/3)^n` or `sum_(n=0)^oo1*(2/3)^n` with the geometric series form `sum_(n=0)^oo a*r^n` , we determine the corresponding values as:

`a=1` and `r= 2/3` .

The `r= 2/3` falls within the condition `|r|lt1` since `|2/3|lt1` or `|0.67| lt1` .

Note: `2/3 ~~0.67` .

By applying the formula: `sum_(n=0)^oo a*r^n= a/(1-r)` , we determine that the given geometric series will converge to a value:

`sum_(n=1)^oo(2/3)*(2/3)^(n -1) =1/(1-2/3)`

`=1/(3/3-2/3)`

`=1/(1/3)`

`=1*(3/1)`

` = 3`

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