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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