HW s 2-3

# HW s 2-3 - MATH 290 WRITING ASSIGNMENT 2 SOLUTIONS 1 Let...

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MATH 290 – WRITING ASSIGNMENT 2 – SOLUTIONS 1. Let { a 1 , a 2 , a 3 , ... } be a sequence of real numbers. Use induction to prove that for all natural numbers n , | a 1 + a 2 + a 3 + ··· + a n | ≤ | a 1 | + | a 2 | + | a 3 | + ··· + | a n | . Solution: The proof will be by induction. If n = 1 then the result reduces to | a 1 | ≤ | a 1 | which is obvious. If n = 2 then the result is | a 1 + a 2 | ≤ | a 1 | + | a 2 | which was proved in Writing Assignment 1. Hence the result holds for n = 1 and n = 2. Let n be a natural number with n > 2 and assume that the result holds for 1 , 2 , ..., n . We must prove that it holds for n + 1. Note that | a 1 + a 2 + a 3 + ··· + a n + a n +1 | = | ( a 1 + a 2 + a 3 + ··· + a n ) + a n +1 | and that since the result holds for n = 2, | ( a 1 + a 2 + a 3 + ··· + a n ) + a n +1 | ≤ | a 1 + a 2 + a 3 + ··· + a n | + | a n +1 | . (Note that this is why we had to establish the base case n = 2 above and not just n = 1.) Since the result holds also for n , | a 1 + a 2 + a 3 + ··· + a n | ≤ | a 1 | + | a 2 | + | a 3 | + ··· + | a

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HW s 2-3 - MATH 290 WRITING ASSIGNMENT 2 SOLUTIONS 1 Let...

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