Question
Asked Nov 26, 2019
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Use the limit comparison test to find the sum from k=1 to infinity of: [(5sin2k)/k!]

 
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Expert Answer

Step 1
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5 sin2 k The given series is k! The domain of sine function is -1,11 This implies, 0 < sin2 ks1.

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Step 2
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.5 sin2 k By applying the Comparison test, 0 s k! 5 k!

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Tagged in

Math

Calculus

Limits