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Determine the value of the series \sum_{k=0}^\infty 5(\frac{3^{k-1}}{2^{2k}})

Question:

Determine the value of the series {eq}\sum_{k=0}^\infty 5(\frac{3^{k-1}}{2^{2k}}) {/eq}

Sum of Series:

We consider the geometrical series

{eq}\sum_{k=0}^\infty r^k = \frac{1}{1-r} {/eq}

The above series is convergent when {eq}|r|<1 {/eq}

The radius of convergence is {eq}r=1 {/eq} and the interval of convergence is {eq}r \in [-1,1] {/eq}

Answer and Explanation:

Given the series

{eq}\sum_{k=0}^\infty 5(\frac{3^{k-1}}{2^{2k}}) {/eq}

we rewrite it as

{eq}\sum_{k=0}^\infty 5(3^{-1})(\frac{3^k}{(2^2)^k})...

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Testing for Convergence & Divergence by Comparing Series

from AP Calculus BC: Exam Prep

Chapter 21 / Lesson 5
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