114 - ∞ X n =3 1 n 2 converges(as it’s a p-series with...

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MATH 153 Selected Solutions for § 11.4 Exercise 12 : Determine whether the series converges or diverges. X n =0 1 + sin n 10 n Solution : Since 0 < 1 + sin n < 2, we have that 1 + sin n 10 n < 2 10 n for all n . Thus, since X n =0 2 10 n converges (it’s a geometric series with r = 1 10 ), by the Comparison Test, our series converges as well. Exercise 14 : Determine whether the series converges or diverges. X n =2 n n - 1 Solution : Note that n n - 1 > n n = 1 n 1 / 2 for every n . Thus, since X n =2 1 n 1 / 2 diverges (it’s a p -series with p < 1), by the Comparison Test our series also diverges. 1
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2 Exercise 22 : Determine whether the series converges or diverges. X n =3 n + 2 ( n + 1) 3 Solution : We would like to compare this to n n 3 = 1 n 2 , but unfortunately we can’t say that n +2 ( n +1) 3 < n ( n 3 , because n + 2 isn’t less than n . Thus, we must use the Limit Comparison Test. lim n →∞ n +2 ( n +1) 3 1 n 2 = lim n →∞ ( n + 2) n 2 ( n + 1) 3 = lim n →∞ n 3 + 2 n 2 n 3 + 3 n 2 + 3 n + 1 = 1 . Since the limit is positive and finite, and since
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Unformatted text preview: ∞ X n =3 1 n 2 converges (as it’s a p-series with p > 1), we know by the Limit Comparison Test that our series converges as well. Exercise 37 : the meaning of the decimal representation of a number 0 .d 1 d 2 d 3 ... (where the digit d i is one of the numbers 0 , 1 , 2 ,... 9) is that .d 1 d 2 d 3 d 4 ... = d 1 10 + d 2 10 + d 3 10 + d 4 10 + ... Show that this series always converges. Solution : Since d i < 10 for all i , we have that .d 1 d 2 d 3 d 4 ... = ∞ X i =1 d i 10 i < ∞ X i =1 10 10 i . This last series is a geometric series with r = 1 / 10, so it converges. Thus, by the Comparison Test, all decimals converge. Note that this is more or less how you can prove that 0 . 9 = 1....
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114 - ∞ X n =3 1 n 2 converges(as it’s a p-series with...

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