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

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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