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If the following series is convergent, find its sum. otherwise, write "div" for divergent.

[infinity]
Σ [3/2ⁿ + (-1/4)ⁿ] =
n=1

User Sash
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1 Answer

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Final answer:

The series ∑ [3/2^n + (-1/4)^n] from n=1 to infinity is convergent. Separating it into two geometric series, the sum is calculated as 6 + 4/5, which equals 34/5 or 6.8.

Step-by-step explanation:

To determine whether the series ∑ [3/2^n + (-1/4)^n] from n=1 to infinity is convergent and to find its sum if it is, we can separate the series into two parts:

  • The sum of the geometric series 3/2^n
  • The sum of the geometric series (-1/4)^n

For the first series, since the common ratio 1/2 is less than 1, the series is convergent and its sum can be found using the formula S = a / (1 - r), where a is the first term and r is the common ratio. So the sum for this part is S1 = 3 / (1 - 1/2) = 6.

For the second series, since the common ratio -1/4 is also less than 1 in absolute value, this series is also convergent. The sum for this part is S2 = 1 / (1 - (-1/4)) = 1 / (1 + 1/4) = 1 / (5/4) = 4/5.

The total sum of the series is then the sum of S1 and S2, which is 6 + 4/5 = 30/5 + 4/5 = 34/5 or 6.8.

If either of the series were divergent, we would simply write "div" for divergent, but both series are convergent in this case.

User Nenu
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