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 nn = + Compare to 2 2 1 n n = using the direct comparison 22 ln 1 ln( n n + , thus 1l n ln( 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( lim lim lim lim lim 1 1 11 ln( 1 n n n n n →∞ →∞ →∞ →∞ →∞ + + ⎛⎞ = = + ⎜⎟ + ⎝⎠ + n = , so again n ln( n n + 2. 2 1 2 n n n n = ! Apply the ratio test.
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