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How do you find the sum of the infinite geometric series #Sigma (0.4)^n# from n=0 to #oo#?

Answer 1

Samantha Adkison

#5/3#

Where #absr<1#, note that #sum_(n=0)^oor^n=1/(1-r)#.

Here, since #abs0.4<1#, we see that

#sum_(n=0)^oo(0.4)^n=1/(1-0.4)=1/0.6=1/(3//5)=5/3#

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Answer 2

Ezekiel Alexander

To find the sum of the infinite geometric series ( \sum_{n=0}^{\infty} (0.4)^n ), you can use the formula for the sum of an infinite geometric series:

[ S = \frac{a}{1 - r} ]

where ( a ) is the first term of the series and ( r ) is the common ratio.

In this series, the first term ( a = (0.4)^0 = 1 ) and the common ratio ( r = 0.4 ).

Plugging these values into the formula:

[ S = \frac{1}{1 - 0.4} = \frac{1}{0.6} = \frac{5}{3} ]

Therefore, the sum of the infinite geometric series ( \sum_{n=0}^{\infty} (0.4)^n ) is ( \frac{5}{3} ) or approximately 1.6667.

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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.