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January 19th, 2019, 04:14 AM   #1
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Solve inequality

Which inequality of the harmonic series should I use to solve the inequality?
$\displaystyle H_n > \frac{3}{n}\; \;$
$\displaystyle H_n =1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}$

Last edited by skipjack; January 19th, 2019 at 05:04 AM.
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January 19th, 2019, 05:09 AM   #2
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As $n$ increases, $H_n$ increases and $3/n$ decreases.
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January 19th, 2019, 09:52 AM   #3
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It is enough to write $\displaystyle \frac{3}{n}=\frac{1}{n}+\frac{1}{n}+\frac{1}{n}$
Write $\displaystyle H_n$ as $\displaystyle \frac{1}{n}+\frac{1}{n-1} +\frac{1}{n-2}+...+1$$\displaystyle >\frac{3}{n}$
The first 3 terms of $\displaystyle H_n$ are larger than $\displaystyle \frac{3}{n}\; \;$, which means the number of terms must be 3 or more (because the number of terms of the H is equal to $\displaystyle n$).
$\displaystyle n\geq 3$
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Last edited by skipjack; January 19th, 2019 at 03:33 PM.
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January 19th, 2019, 02:19 PM   #4
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Note that the inequality holds only for $n\ge 3$.
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January 19th, 2019, 02:43 PM   #5
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I meant inequation; I wrote it wrong.
Not inequality but inequation.

Last edited by skipjack; January 19th, 2019 at 03:34 PM.
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January 20th, 2019, 12:56 PM   #6
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Quote:
Originally Posted by idontknow View Post
I meant inequation; I wrote it wrong.
Not inequality but inequation.
What do you mean by "inequation"? I don't believe there is such a word.
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January 20th, 2019, 01:00 PM   #7
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I mean: find interval of $\displaystyle n$ or the range of $\displaystyle n$
Such that inequality $\displaystyle H_n \geq \frac{n}{3} $ holds true.
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January 20th, 2019, 01:54 PM   #8
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Are you sure you mean n/3 rather than 3/n?
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January 20th, 2019, 02:32 PM   #9
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3/n .
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