How do you find the sum of the infinite geometric series #Sigma -3(0.9)^n# from n=0 to #oo#?
# S = sum_(n=0)^oo -3(0.9)^n = -30#
The sum:
Which gives us:
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To find the sum of the infinite geometric series ( \sum_{n=0}^\infty -3(0.9)^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
- ( r ) is the common ratio
In this series, ( a = -3 ) and ( r = 0.9 ). Substituting these values into the formula:
[ S = \frac{-3}{1 - 0.9} ]
[ S = \frac{-3}{0.1} ]
[ S = -30 ]
So, the sum of the infinite geometric series ( \sum_{n=0}^\infty -3(0.9)^n ) is -30.
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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.
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.
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.
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.
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