Convergence of a geometric series; rewriting a series in the form ar^(n-1)

[lilypetals]

Homework Statement

Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum.
$$\sum$$_n=1^infinity (-3)^n-1/4ⁿ

Homework Equations

A geometric series, $$\sum$$_n=1^infinity ar^n-1=a + ar + ar² + ... is convergent if |r|< 1 and its sum is $$\sum$$_n=1^infinity ar^n-1 = a/(1-r), |r| < 1. If |r| $$\geq$$ to 1, the geometric series is divergent.

The Attempt at a Solution

I know that I need to rearrange the series to reflect ar^n-1, but I'm not sure how to go about doing that. Any suggestions?

 

[lilypetals]

Actually, I just found a (surprisingly) helpful hint in the small-print margin of my textbook: we write out the first few terms to determine a and r of the series.

a₁=1/4
a₂=-3/16
a₃=9/64

So the series becomes 1/4(-3/4)^n-1, which is convergent, because r=-3/4, which is less than 1.

And its sum is equal to a/(1-r) = (1/4)/(1--3/4) = (1/4) * (4/7) = 1/7.

Is this correct?

 

[Dick]

Sure, that's right.