January 19th, 2019, 04:14 AM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68  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. 
January 19th, 2019, 05:09 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
As $n$ increases, $H_n$ increases and $3/n$ decreases.

January 19th, 2019, 09:52 AM  #3 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
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}{n1} +\frac{1}{n2}+...+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 $\displaystyle n$). $\displaystyle n\geq 3$ Last edited by skipjack; January 19th, 2019 at 03:33 PM. 
January 19th, 2019, 02:19 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670 
Note that the inequality holds only for $n\ge 3$.

January 19th, 2019, 02:43 PM  #5 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
I meant inequation; I wrote it wrong. Not inequality but inequation. Last edited by skipjack; January 19th, 2019 at 03:34 PM. 
January 20th, 2019, 12:56 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,704 Thanks: 670  
January 20th, 2019, 01:00 PM  #7 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
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. 
January 20th, 2019, 01:54 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
Are you sure you mean n/3 rather than 3/n?

January 20th, 2019, 02:32 PM  #9 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
3/n .


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