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Solutions cal 2 test 2

# Solutions cal 2 test 2 - Sample Test 2 Solutions 1 Do the...

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Sample Test 2 Solutions 1. Do the following converge (explain)? (1.1) n = 1 ln n n 4 + 1 , Compare with n = 1 n n 4 + 1 . Since ln n < n for n 1, then ln n n 4 + 1 < n n 4 + 1 . This implies that n = 1 ln n n 4 + 1 < n = 1 n n 4 + 1 . Since n = 1 n n 4 + 1 converges (direct comparison with n = 1 1 n 3 p = 3 ) then by the direct comparison test (DCT), the original series converges. (1.2) n = 1 1 n 3 + 1 , Compare with n = 1 1 n 3 . Since n = 1 1 n 3 converges ( p series with p = 3) then by the limit comparison test (LCT), the original series converges. (1.3) n = 1 1 2 + 1 n · n , Taking the limit lim n n p | a n | = lim n 1 2 + 1 n · = 1 2 < 1 then by the n th root test, the original series converges. (1.4) n = 1 e n n ! , Consider lim n fl fl fl a n + 1 a n fl fl fl = lim n e n + 1 ( n + 1 ) ! / e n ( n ) ! = lim n e n + 1 ( n + 1 ) ! · ( n ) ! e n = lim n e n + 1 = 0 < 1 so by ratio test, the series converges (1.5) n = 1 1 ln ( n + 1 ) , Since ln ( n + 1 ) < n + 1 for n 1 then 1 n + 1 < 1 ln ( n + 1 ) for n 1 and since n = 1 1 ( n + 1 ) (harmonic) diverges, then by the DCT, original series does as well. (1.6) n = 1 1 n ( n + 1 ) , Comparing with n = 1 1 n 2 then lim n 1 n ( n + 1 ) / 1 n 2 = lim n n 2 n ( n + 1 ) = 1, and since n = 1 1 n 2 converges (p-series with p = 2) then by the limit comparison test (LCT) the original series converges. (1.7) n = 1 n - 1 n + 1 , Since lim n n - 1 n + 1 = 1, then by the n th term test for divergence, the series diverges. (1.8) n = 1 ( 2 n ) ! ( n ! ) 2 , 1

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Consider lim n fl fl fl a n + 1 a n fl fl fl = lim n ( 2 n + 2 ) ! ( n + 1 ) ! 2 / ( 2 n ) ! ( n ) ! 2 = lim n ( 2 n + 2 ) ! ( 2 n ) ! · ( n + 1 ) ! 2 n ! 2 = lim n ( 2 n + 2 )( 2 n + 1 ) ( n + 1 )( n + 1 ) = 4 > 1 so by ratio test, the series diverges (1.9) n = 2 1 ln 2 ( n ) , Since ln n < n for n 1 then ln 2 n < n ln n for n 1 which gives 1 n ln n < 1 ln 2 n for n 1. Since n = 1 1 n ln n diverges, (see next question) then by the direct comparison test, original series does as well.
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