How do you find the sum of the infinite series #Sigma2(1/10)^k# from k=1 to #oo#?

Answer 1

#sum_(n=1)^oo 2(1/10)^n = 2/9#

Start from the geometric series of ratio #r=1/10#:
#sum_(n=0)^oo (1/10)^n = 1/(1-1/10) = 10/9#
Extract from the sum the term for #n= 0#:
#1+ sum_(n=1)^oo (1/10)^n =10/9#
#sum_(n=1)^oo (1/10)^n =10/9-1 = 1/9#
Multiplying by #2# term by term:
#sum_(n=1)^oo 2(1/10)^n = 2 sum_(n=1)^oo (1/10)^n = 2/9#
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Answer 2

To find the sum of the infinite series (\sum_{k=1}^{\infty} 2 \left(\frac{1}{10}\right)^k), you can use the formula for the sum of an infinite geometric series.

The formula is: [ S = \frac{a}{1 - r} ]

Where: ( S ) = sum of the series ( a ) = first term of the series ( r ) = common ratio of the series

In your series, ( a = 2 ) and ( r = \frac{1}{10} ). Plug these values into the formula and solve for ( S ):

[ S = \frac{2}{1 - \frac{1}{10}} ]

[ S = \frac{2}{\frac{9}{10}} ]

[ S = 2 \times \frac{10}{9} ]

[ S = \frac{20}{9} ]

So, the sum of the infinite series (\sum_{k=1}^{\infty} 2 \left(\frac{1}{10}\right)^k) is (\frac{20}{9}).

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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.

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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