How do you find the sum of the infinite geometric series #Sigma (1/2)^n# from n=0 to #oo#?

Answer 1

The sum is 2.

Realize that the sum of a geometric series of the form #sum ar^n# can be represented by #a/(1-r)# where #a# is the first term of the series and #r# is the common ratio. Thus we can see that the series #sum (1/2)^n# is of the form of a geometric series, where the r is 0.5 and the a is 1.

So the sum of our series becomes #1/(1-(1/2))# which is #1/0.5 = 2#

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Answer 2

Emilia Aguilar

To find the sum of the infinite geometric series ( \sum_{n=0}^\infty \left(\frac{1}{2}\right)^n ), you can use the formula for the sum of an infinite geometric series, which is given by ( \frac{a}{1 - r} ), where ( a ) is the first term of the series and ( r ) is the common ratio.

In this series, ( a = 1 ) and ( r = \frac{1}{2} ). Plugging these values into the formula, you get:

[ \text{Sum} = \frac{1}{1 - \frac{1}{2}} = \frac{1}{\frac{1}{2}} = 2 ]

So, the sum of the infinite geometric series ( \sum_{n=0}^\infty \left(\frac{1}{2}\right)^n ) is ( 2 ).

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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.