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hwsol[1]

# hwsol[1] - lim n!1 6& 1 n ± 1 = 1 7.4.16 P 1 n =1 1(1...

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Solutions to 4/15 written hw 7.3.10 P 1 n =1 ln n n diverges by Direct Comparison with the divergent p series P 1 n : : ln n n > 1 n for n > e: You could also use the Limit Comparison Test. lim n !1 (ln n ) =n 1 =n = lim n !1 ln n = 1 ; so you have a BIGGER THAN BIG situation. 7.3.20 P 1 n =1 1 (ln 3) n is a convergent geometric series with r = 1 ln 3 < 1 : 7.3.26 P 1 n =1 2 1+ e n converges by Direct Comparison with the convergent geometric series P 1 n =1 2 e n : 2 1 + e n < 2 e n : You could also apply the Ratio Test to get ° = 1 =e: Another way to solve this problem would be to apply the LCT using P 1 e n as a benchmark. 7.3.28 P 1 n =1 n n 2 +1 diverges by the LCT with using P 1 n as a benchmark: lim n !1 n n 2 +1 1 n = lim n !1 n 2 n 2 + 1 = 1 : Hence both series diverge. You can also use a direct comparison: n n 2 + 1 > n n 2 + n 2 = 1 2 n : This gives a BIGGER THAN BIG comparison. 7.4.8 P 1 n =1 1 p n 3 +1 converges by Direct Comparison with the convergent p -series P 1 n 3 = 2 : 1 p n 3 + 1 < 1 p n 3 = 1 n 3 = 2 : This gives a SMALLER THAN SMALL comparison. 7.4.12 P 1 n =1 (ln n ) 3 n 3 converges by the LCT with the convergent p ° series P 1 n 2 : Using L°Hopital°s Rule, we see lim n !1 (ln n ) 3 n 3 1 n 2 = lim n !1 (ln n ) 3 n = lim n !1 3 (ln n ) 2 1 n 1 = lim n !1 3(ln n ) 2 n = lim n !1 6(ln n ) 1 n 1 = lim n !1 6 ln n n = lim

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Unformatted text preview: lim n !1 6 & 1 n ± 1 = 1 7.4.16 P 1 n =1 1 (1+ln n )) 2 diverges by the Limit Comparison Test using the divergent p & series P 1 =n : lim n !1 a n b n = lim n !1 n (1 + ln n ) 2 = lim n !1 1 2 (1 + ln n ) ± 1 n = lim n !1 n 2 + 2 ln n = lim n !1 1 2 n = 1 : Alternatively, you can show divergence using Direct Comparison: 1 (1 + ln n ) 2 > 1 (2 p n ) 2 = 1 4 n : This gives a BIGGER THAN BIG comparison. 7.5.2 P n 2 e n converges by the Ratio Test: & = lim n !1 ( n + 1) 2 e n +1 ± e n n 2 = lim n !1 ( n + 1) 2 n 2 ± 1 e = 1 e < 1 : 7.5.12 P (ln n ) n n n converges by the Root Test: & = lim n !1 & (ln n ) n n n ± 1 =n = lim n !1 ln n n = 0 < 1 : 7.5.16 P n ln n 2 n convergest by the Ratio Test: & = lim n !1 ( n + 1) ln ( n + 1) 2 n +1 ± 2 n n ln n = lim n !1 n + 1 n ± ln ( n + 1) ln n ± 1 2 = 1 2 < 1 : 2...
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hwsol[1] - lim n!1 6& 1 n ± 1 = 1 7.4.16 P 1 n =1 1(1...

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