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 January 5th, 2017, 05:42 AM #1 Member   Joined: Oct 2012 Posts: 57 Thanks: 0 Show that Show that : 1+1/2+1/3+....+1/n>2n/(n+1) for n>1
 January 5th, 2017, 01:30 PM #2 Global Moderator   Joined: May 2007 Posts: 6,343 Thanks: 534 mathematical induction works. 1+1/2+...+1/n+1/(n+1)>2n/(n+1)+1/(n+1)=(2n+1)/(n+1) Need to show: (2n+1)/(n+1)>2(n+1)/(n+2) or $\displaystyle (2n+1)(n+2)>2(n+1)^2$ or $\displaystyle 2n^2+5n+2>2n^2+4n+2$. Thanks from fahad nasir
 January 5th, 2017, 03:16 PM #3 Member   Joined: Dec 2016 From: USA Posts: 46 Thanks: 11 It's a routine induction proof. Verify the base case, $n = 2$. Assume it's true for some positive integer $n > 1$. Then, using that assumption, show that the claim holds for the "$n+1$" case. Hint: Start by adding $1/(n+1)$ to both sides.
January 16th, 2017, 09:15 PM   #4
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Quote:
 Originally Posted by fahad nasir Show that : 1+1/2+1/3+....+1/n>2n/(n+1) for n>1
I use Gauss's idea that he use when he solves the problem '1+2+...+n+...+100'.

A useful inequality: with $a, b>0$, we obtain
$$\frac{1}{a}+\frac{1}{b} \ge \frac{4}{a+b}.$$

Therefore,
$$\frac{1}{k}+\frac{1}{n+1-k} \ge \frac{4}{n+1}\,\forall k=1, 2, ..., n.$$
Hence,
$2\sum_{k=1}^{n}\frac{1}{k} \ge \frac{4n}{n+1}.$
...

Note that: with $a, b>0$,
Since
$\frac{1}{a}+\frac{1}{b} = \frac{4}{a+b} \iff a=b.$

Last edited by The Epsilon; January 16th, 2017 at 09:39 PM.

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