What is the sum of the infinite geometric series #sum_(n=1)^oo(1/3)^n# ?

Answer 1

This is a geometric series with first term #1/3# and common ratio #1/3# (see geometric series ). The common ratio is less than 1 in absolute value, so that the series is convergent and its sum is given by #"first term"/(1-"common ratio")=(1/3)/(1-(1/3))=1/2#

You need to review geometric series Geometric Series

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Answer 2

Aurora Alvarado

The sum of the infinite geometric series ( \sum_{n=1}^\infty \left(\frac{1}{3}\right)^n ) can be calculated using the formula for the sum of an infinite geometric series, which is ( \frac{a}{1 - r} ), where ( a ) is the first term and ( r ) is the common ratio.

In this series, the first term ( a ) is ( \left(\frac{1}{3}\right)^1 = \frac{1}{3} ) and the common ratio ( r ) is ( \frac{1}{3} ).

Substituting these values into the formula:

[ \text{Sum} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} ]

So, the sum of the infinite geometric series ( \sum_{n=1}^\infty \left(\frac{1}{3}\right)^n ) is ( \frac{1}{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.