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What is the sum of the series infinity e n-1 -3(3/8)^n?

What is the sum of the series infinity e n-1 -3(3/8)^n?-example-1

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10 votes

the answer for the question above is 9/5

User Cmbaxter
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9 votes

The calculated sum of the series
\sum\limits^(\infty)_(n=1) -3((3)/(8))^n to infinity is (a) -9/5

How to determine the sum of the series to infinity

From the question, we have the following parameters that can be used in our computation:


\sum\limits^(\infty)_(n=1) -3((3)/(8))^n

Set n = 1 to calculate the first term, a

So, we have

a = -3 * 3/8

a = -9/8

The common ratio is the expression in bracket

So, we have

r = 3/8

The sum of the series to infinity is calculated as

Sum = a/(1 - r)

Substitute the known values into the equation

Sum = -9/8/(1 - 3/8)

Sum = -9/8/(5/8)

Evaluate

Sum = -9/5

Hence, the sum of the series to infinity is (a) -9/5

User Jonathan Robbins
by
7.9k points

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