Question

How do you test `\sum _ { m = 1} ^ { \infty } \frac { ( - 6) ^ { m + 1} } { 4^ { 8m } }` for convergence or divergence?

Answer 1

This series is convergent by the geometric series test.

Explanation:

`sum_(m=1)^oo frac{(-6)^(m+1)}{4^(8m)}`

Rewrite the fraction using exponent rules
`sum_(m=1)^oo frac{(-6)^(m)(-6)}{32^(m)}`

`sum_(m=1)^(oo) (-6)((-6)/32)^m`

`sum_(m=1)^(oo) -6((-3)/16)^m`

Because the absolute value of the common ratio `|r| = |(-3)/16| = 3/16`is less than `1`, the series is convergent by the geometric series test.

To find what it converges to:
Geometric series that converge converge to `a/(1-r)` where `a` is the first term of the series, and `r` is the common ratio.

`a/(1-r) = frac{-6(-3/16)}{1+ 3/16}`

`= ((9/8))/((19/16))`

`= 9/8 * 16/19`

`= 18/19`