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Asked Nov 26, 2019
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Expert Answer

Step 1

We will be using limit comparison test to determine if the series converges,

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Σ 214 466n11 106 n 1 For the limit comparison test, we will be following the steps as (A), (B) and (C)

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Step 2

For [A],

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214 1 Notice that for larger value of n 466n11-106 n11/6 So, when performing the limit comparison test, the best series to compare with would 1 be 1bhere bn n=1Dn n11/6

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Step 3

For [B],

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214 аn lim 466п11-106 lim 1 n0o bn пэ00 ni1/6 'n11/6 X 214 = lim V466n11-106 пз0о = lim 214n11/6 V466n11-106. n0o 214 lim 6 466n11-106 214 = lim 6 (466n11-106) пзо nli 214 = lim 6 106 пз0о 466 n1 214 V466 = 76.857 > 0 аn lim 76.857 So, пэ00 bn

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