`sum_(n=0)^oo 4(-1.05)^n` Verify that the infinite series diverges
Recall that an infinite series converges 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 ` .
The given infinite series `sum_(n=0)^oo 4(-1.05)^n` resembles the form of geometric series with an index shift: `sum_(n=0)^oo a*r^n` .
By comparing "`4(-1.05)^n` " with "`a*r^n` ", we determine the corresponding values: `a = 4` and `r =-1.05` .
The convergence test for the geometric series follows the conditions:
a) If `|r|lt1` or `-1 ltrlt1 ` then the geometric series converges to `sum_(n=0)^oo a*r^n = a/(1-r)` .
b) If `|r|gt=1` then the geometric series diverges.
The `r=-1.05` from the given infinite series falls within the condition `|r|gt=1` since `|-1.05|gt=1` . Therefore, we may conclude that infinite series `sum_(n=0)^oo 4(-1.05)^n` is a divergent series.