Exponential Series ProofDate: 05/05/2001 at 22:55:22 From: Jake Subject: Exponents/Algebra Hi, I'm having trouble with an exponential question. In previous parts of the question, they give it that e^x is greater than or equal to 1 + x for all real values of x. They also tell you that: (1+1)(1+(1/2))(1+(1/3)) ... (1+(1/n)) = n+1 The question is: Using these two pieces of information, prove that: e^(1 + (1/2) + (1/3) + ... + (1/n)) > n and also, Find a value of n for which: 1 = (1/2) + (1/3) + ... + (1/n) > 100 Thanks! Date: 05/06/2001 at 17:18:24 From: Doctor Jaffee Subject: Re: Exponents/Algebra Hi Jake, Here is how I would prove that e^(1 + (1/2) + (1/3) + ... + (1/n)) > n. Since e^x is greater than or equal to 1 + x, it follows that: e^1 is greater than or equal to 1 + 1, e^(1/2) is greater than or equal to 1 + 1/2, e^(1/3) is greater than or equal to 1 + 1/3 . . . e^(1/n) is greater than or equal to 1 + 1/n. So, the product of (e^1)(e^(1/2))...(e^(1/n)) is greater than (1 + 1)(1 + 1/2)(1 + 1/3)...(1 + 1/n) Now, you can use the rule for multiplying exponential numbers to transform the left side into what you want, and the right side you know equals n + 1. You should be able to work it from there. I'll give you a hint for the second problem. You know that 1/3 is greater than 1/4, so 1/3 + 1/4 must be greater than 2/4 or 1/2. Likewise, 1/5, 1/6, and 1/7 are all greater than 1/8, so the sum of the four of them must be greater than 4/8 or 1/2. In a like fashion, the sum of the next eight fractions is greater than 1/2, the sum of the sixteen fractions that follow is greater than 1/2, etc. You should be able to see how many 1/2's you need to exceed 100. Give these problems a try and if you want to check your answers with me, write back. If you are having difficulties, let me know and show me what you have done so far, and I'll try to help you some more. Good luck. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/ Date: 05/11/2001 at 04:42:58 From: Jake Subject: Re: Exponents/Algebra Thank you so much! It helped a lot. About the second problem, I understand your instructions but could you do it like this: They ask you to find 1+ 1/2 + 1/3 +...+1/n > 100. Since they tell you that e^(1+1/2+1/3+...1/n) > 100 in the previous question, does it make sense just to say that it implies that any n > e^100 will do? From: Doctor Jaffee Subject: Re: Exponents/Algebra Hi, Yes, that makes perfectly good sense to me. What I wrote you before was correct, but it occurred to me later that a better approach would be to take the natural logarithm of both sides of e^(1+(1/2)+(1/3)+...+(1/n)) > n then set ln n = 100. I assume that you ended up doing something similar to that to arrive at your conclusion. I'm glad I was able to help you; I enjoyed working on the problem and I hope to receive more interesting questions from you. - Doctor Jaffee, The Math Forum http://mathforum.org/dr.math/ |
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