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How do you find the sum of the infinite geometric series #Sigma 4(1/4)^n# from n=0 to #oo#?

Answer 1

Christopher Allers

#Sigma_0^oo 4(1/4)^n=5 1/3#

The first term of the series #Sigma_0^oo 4(1/4)^n# is

#4(1/4)^0=4xx1=4# and as each successive term #1/4# times its immediately preceding term common ratio is #1/4#.

and let the sum of the series #S_oo# be #4+1+1/4+1/16+1/64+...........#

As #S_oo=4+1+1/4+1/16+1/64+...........#, we have

#4S_oo=16+4+1+1/4+1/16+1/64+...........#

Now subtracting first from second we get

#3S_oo=16# and #S_oo=16/3=5 1/3#

Note - In a geometric series with #a# as first term and common ratio as #r#, #S_oo=a/(1-r)#, hence sum of series is #4/(1-1/4)=4/(3/4)=4xx4/3=16/3=5 1/3#.

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Answer 2

Aurora Alvarado

The sum of the infinite geometric series Σ 4(1/4)^n from n=0 to ∞ is equal to 4/(1 - 1/4), which simplifies to 4/(3/4) = 16/3.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.