`sum_(n=0)^oo 4(-1.05)^n` Verify that the infinite series diverges

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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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