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18.100pset5

# 18.100pset5 - 18.100B/C Fall 2008 Solutions to Homework 5...

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18.100B/C: Fall 2008 Solutions to Homework 5 Every use of the phrase “by comparison” in the following solutions is a reference to Rudin’s 3.25, the comparison test. 1.(a) We have 0 < a n , so 0 < a n 2 + n 2 a n = 1 n 2 + 2 /a n < 1 n 2 . Thus n =1 a n 2+ n 2 a n converges by comparison with the convergent series n =1 1 n 2 . (b) Following the hint, we first suppose a n 0. Then there is some r > 0 such that a n > r for infinitely many values of n . But then a n 2+ a n = 1 1+ 2 an > 1 1+ 2 r = r 2+ r > 0 for infinitely many n so a n 2+ a n 0 and thus n =1 a n 2+ a n diverges (Rudin 3.23). Now instead suppose a n 0. Choose some N such that for all n N we have a n < 1. Then we have a n a n +2 > a n 3 , n =1 a n a n +2 diverges by comparison with 1 3 n =1 a n . 2.(a) The first part of the problem is an ugly-looking computation. Here’s a short version (feel free to add your own summation signs): the left side contains terms of the form a 2 i b 2 j for i, j ∈ { 1 , . . . , n } , and each term appears exactly once. The first term on the left contains terms of the form a 2 i b 2 i , each appearing once, and terms of the form a i b i a j b j for i = j , each appearing once. The second term on the left is the complicated part: for i = j we get one-half copy of a 2 i b 2 j , one half copy of a 2 j b 2 i and one negative copy of a i b j a j b i , while for i = j we get no contribution. Observe that summing one-half copy of a 2 i b 2 j and one half copy of a 2 j b 2 i over all pairs i = j gives you the same thing as summing one copy of a 2 i b 2 j over all pairs i = j , and celebrate. (Note that this argument glosses over the formal justification for its claims. In particular, a more careful solution-writer would have explicitly used an inductive proof.) Since squares of real numbers are nonnegative, it follows immediately that n i =1 a i b i 2 n i =1 a 2 i

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