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{(1/sqrt(2))^n} from n=0 to infinity; is the series convergent or divergent? if convergent, why? and sum?
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{(1/sqrt(2))^n} from n=0 to infinity; is the series convergent or divergent? if convergent, why? and sum?
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Jun 27, 2017
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{(1/sqrt(2))^n} from n=0 to infinity; is the series convergent or divergent? if convergent, why? and sum?
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Yorkw
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The series is convergent.
Step-by-step explanation:
1/∞ = 0 ( it would be divergent if the answer was: -∞ or +∞ )
Sum of the infinite geometric series:
S∞ = a1 / ( 1-r )
a1 = 1 a2=
, a3=1/2, a4=
...
r =
S∞ =
Brennan Cheung
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Jul 3, 2017
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Brennan Cheung
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