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.

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