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

1 Answer
Anjali G
May 20, 2017

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#

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