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133quiz8solution - Quiz 8 Name Section TA Test for...

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Quiz 8 Name____________ Section___________ TA______________ Test for convergence or divergence. 1. 2 2 ln ln( 1) n n n n = + Compare to 2 2 1 n n = using the direct comparison 2 2 ln 1 ln( 1) n n n n + , thus 2 2 2 2 1 ln ln( 1) n n n n n n = = < ∞ < ∞ + . Alternatively, we could use the limit comparison test, comparing again to 2 2 1 n n = : 2 2 ln 1 ln 1 1 ln( 1) lim lim lim lim lim 1 1 1 1 ln( 1) 1 n n n n n n n n n n n n n n n →∞ →∞ →∞ →∞ →∞ + + = = = = + + + n = , so again 2 2 2 2 1 ln ln( 1) n n n n n n = = < ∞ < ∞
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