1873_solutions

Theorem a given that f is an integrable function on

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Unformatted text preview: nclude that np lim nÝ jn1 1  log p. j log j Some Exercises on The Comparison Test Test each of the following series for convergence. 1 n 3/2  n 1. Solution: We define an  1 n 3/2  n and 1 n 3/2 b n is a convergent p-series and since bn  for each n. Since the series 1 a n  lim n 3/2  n  lim n 3/2 lim n Ý bn nÝ n Ý n 3/2  n 1 n 3/2 1 nÝ lim 1 1  1n we see that a n is convergent. Note that we could have achieved this solution more rapidly by observing that a n b n for each n. However, taking the limit was worth while because it applies just as well to Exercise 2. 2. 1 n 3/2 n We define an  1 n 3/2 n and 1 n 3/2 for each n. Since b n is a convergent p-series and since lim a n  1 n Ý bn a n is convergent. we deduce that bn  3. n4 n n2  2 Solution: In order to see how to proceed observe that, for large n, the expression n4 n n2  2 behaves like n 1 n n4 because n 2  2 is much smaller than n 4 . Now we begin. 324 We define an  n4 n n2  2 and for each n. Since we deduce that bn  1 n b n is a divergent p-series and since n 4 n2  2 n lim a n  n Ý lim nÝ lim 1 n Ý bn n n4 n4 1 n2  2 a n is divergent. n log n 4. n2  2 n5 Solution: The key to this exercise is the fact that, although n log n is larger than n, it is not much larger because of the fact that log n 0 n lim nÝ whenever   0. The denominator of the fraction n log n n2  2 n5 behaves like n 5/2 . We can therefore consider that n log n n2  2 n5 behaves like log n n log n  3/2  n n 5/2 log n n 1/4 1 n 5/4 . With this in mind, we define n log n an  n5 n2  2 and 1 n 5/4 b n is a convergent p-series and that log n lim a n  n Ý 1/4  0 lim n Ý bn n bn  for each n. We observe that a n is convergent. and so 5. 1 n 1 log n /n Solution: We recall that if a and b are any positive numbers then a b  exp b log a . We begin by observing that if n is any positive integer then log n log n n 1 log n /n  n n log n /n  n exp n Since 325  n exp log n n 2 log n n lim exp nÝ 2 log n n  exp n Ý lim 2  exp 0  1 we conclude that 1 1 log n /n n behaves like 1/n for large n . We define 1 an  n 1 log n /n and for each n. Since bn  1 n b n is a divergent p-series and since 1 lim a n  n Ý lim n Ý bn log n exp 1 2 n we conclude that a n is divergent. 1 6. n 1 log n 2 /n Hint: This exercise is similar to Exericse 5 because of the identity n 1 log n 2 /n 3 log n n  n exp and because log n n lim exp nÝ 1 Thus n 3 log n n  exp n Ý lim 3  exp 0  1. is divergent. 1 log n 2 /n 1 7. n 1 log n log log n /n Solution: For each n we see that n 1 log n log log n  n exp /n log n log n n log log n Now log n log log n  exp log log n 2 log log n log n  exp 2 and since lim nÝ log log n log n 2 0 we conclude that if n is sufficiently large then log n log log n  exp log log n log n  exp 1 log n 2 Therefore, for n sufficiently large we have log n log n 0 n log log n  and we conclude that 326 2 log n  n 1/2 log n n 1/2 log n  1/2 n n log n log n lim nÝ log log n...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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