# `sum_(n=1)^oo 2(-1/2)^n` Verify that the infinite series converges

To verify if the given infinite series: `sum_(n=1)^oo 2(-1/2)^n` converges, recall that infinite series converge to a single finite value S  if the limit of the partial sum `S_n ` as n approaches ` oo` converges to `S` . We follow it in a formula:

`lim_(n-gtoo) S_n=sum_(n=1)^oo a_n = S` .

To evaluate the  `sum_(n=1)^oo 2(-1/2)^n` , we apply the Law of exponent : `x^(n+m) = x^n*x^m` .

Then, `(-1/2)^n =(-1/2)^(n -1+1)`

`=(-1/2)^(n -1)*(-1/2)^1 `

` = (-1/2)^(n -1)*(-1/2)`

Plug-in `(-1/2)^n =(-1/2)^(n -1)*(-1/2)` , we get:

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

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

By comparing given infinite series  `sum_(n=1)^oo -1*(-1/2)^(n -1)` with the geometric series form `sum_(n=1)^oo a*r^(n-1)` , we determine the corresponding values as:

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

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.

The `r=-1/2` falls within the condition` |r|lt1` since `|-1/2|lt1` or `|-0.5| lt1` .

Therefore, the series converges.

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

`sum_(n=1)^oo2(-1/2)^n=sum_(n=1)^oo -1*(-1/2)^(n -1)`

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

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

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

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

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

` = -2/3 or -0.67 ` (approximated value)

Approved by eNotes Editorial Team