Unformatted text preview: 2. Prove that 1 / âˆš 1 + 1 / âˆš 2 + Â·Â·Â· + 1 / âˆš n â‰¥ âˆš n for all n âˆˆ N . We, again, argue using induction. The n = 1 base case follows trivially as both sides are equal to 1 in that case. We then assume that the result holds for n replaced by k1 and seek to use that to prove the n = k case. Here, by this inductive hypothesis, we have 1 / âˆš 1 + 1 / âˆš 2 + Â·Â·Â· + 1 / âˆš k1 + 1 / âˆš k â‰¥ âˆš k1 + 1 âˆš k . Thus, it will suï¬ƒce to show that âˆš k1 + 1 âˆš k â‰¥ âˆš k. Note that p k ( k1) â‰¥ p ( k1) 2 = k1 , for k â‰¥ 2 . Thus, p k ( k1) + 1 â‰¥ k in this range. Or âˆš k1 + 1 âˆš k â‰¥ âˆš k, which is the desired inequality. 1...
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 Spring '10
 MetCalfe
 Calculus, Logic, Inductive Reasoning, Negative and nonnegative numbers, Equals sign

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