My Math Forum Proof of divergence for the harmonic series

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 April 16th, 2012, 06:26 PM #1 Newbie   Joined: Apr 2012 Posts: 17 Thanks: 0 Proof of divergence for the harmonic series Book says that step one is to show that for n > 1 $\frac{1}{n-1} + \frac{1}{n} + \frac{1}{n+1} > \frac{3}{n}$ I know how to algebraic show that this is the same as $\frac{2n^2}{n^2-1} > 2$ And this is the same as (1) $\frac{2n^2}{(n-1)(n+1)} > 2$ Now the book shows in the solution that the next step is (2) $\frac{n}{n-1} + \frac{n}{n+1} > 2$ How did they go from (1) to (2) algebraically? I thought it would be by using the method of partial fractions but when I did that I obtained $\frac{1}{n-1} + \frac{-1}{n+1}$which is not correct.
 April 16th, 2012, 06:33 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Proof of divergence for the harmonic series You have made an error in your partial fraction decomposition. You could get to (2) by beginning with: $\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}>\frac{3}{n}$ Multiply through by n $\frac{n}{n-1}+1+\frac{n}{n+1}>3$ (2) $\frac{n}{n-1}+\frac{n}{n+1}>2$
 April 16th, 2012, 06:38 PM #3 Newbie   Joined: Apr 2012 Posts: 17 Thanks: 0 Re: Proof of divergence for the harmonic series That makes perfect sense but why in the heck did they bother mentioning the first step to be $\frac{2n^2}{n^2-1} > 2$? Then they went to (2). I just don't understand why they did it this way instead of your method. Do you know why?
 April 16th, 2012, 06:41 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Proof of divergence for the harmonic series No, I couldn't say, but maybe they are going to use (1) at some point later in the proof. I doubt that was done unnecessarily.
 April 16th, 2012, 06:47 PM #5 Newbie   Joined: Apr 2012 Posts: 17 Thanks: 0 Re: Proof of divergence for the harmonic series So does what you have shown prove that for n > 1 $\frac{1}{n-1} + \frac{1}{n} + \frac{n+1} > \frac{3}{n}$? If so, why does it prove it? I don't understand how solving for n proves the statement above.
 April 16th, 2012, 06:49 PM #6 Newbie   Joined: Apr 2012 Posts: 17 Thanks: 0 Re: Proof of divergence for the harmonic series Should be $\frac{1}{n-1} + \frac{1}{n} + \frac{1}{n+1} > \frac{3}{n}$ and I did not mean solve for n. I meant having n on one side of the equation.
 April 16th, 2012, 07:00 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Proof of divergence for the harmonic series If I were going to prove the original statement, I would write it in the form: $2+\frac{2}{n^2-1}>2$ $\frac{2}{n^2-1}>0$ We know this is true for any finite value of 1 < n.
 April 17th, 2012, 11:02 AM #8 Newbie   Joined: Apr 2012 Posts: 17 Thanks: 0 Re: Proof of divergence for the harmonic series Ok then next part of the question says: By grouping the terms of the series by threes, verify the inequality 1 + (1/2 + 1/3 + 1/4) + (1/5 + 1/6 + 1/7) + ... > 1 + 3/3 + (3/6 + 3/9 + 3/12) + ... > 1 + 3/3 + 9/9 + 27/27 + ... How do I even begin to verify the inequality? Also, I am confused on what exactly it is asking me to verify. I am guessing it wants me to verify that it holds for any n using induction?
 April 17th, 2012, 11:46 AM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Proof of divergence for the harmonic series We have verified: $\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}>\frac{3}{n}$ Hence we know: $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>\frac{3}{3}$ $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}>\frac{3}{6}$ $\frac{1}{8}+\frac{1}{9}+\frac{1}{10}>\frac{3}{9}$ etc. Now, since $\frac{1}{n-1}+\frac{1}{n}+\frac{1}{n+1}>\frac{3}{n}$ we also have: $\frac{3}{3(n-1)}+\frac{3}{3n}+\frac{3}{3(n+1)}>\frac{9}{3n}$ which is required for the last step in the given inequality.
April 18th, 2012, 10:33 AM   #10
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Re: Proof of divergence for the harmonic series

Quote:
 Originally Posted by MarkFL If I were going to prove the original statement, I would write it in the form: $2+\frac{2}{n^2-1}>2$ $\frac{2}{n^2-1}>0$
I did you go from our original statement to this? I thought I understood it yesterday but I cannot seem to manipulate our given information into this.

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