Induction Proof (Inequality) SIMPLE!!

**GamesonPlanes** · March 2nd 2010, 04:52 PM

Show 1/(n+1)+1/(n+2)+1/(n+3)+1/2n>3/5, for n>2. Use mathematical induction.

THANKS

**Archie Meade** · March 2nd 2010, 05:13 PM

Originally Posted by GamesonPlanes

Show 1/(n+1)+1/(n+2)+1/(n+3)+1/2n>3/5, for n>2. Use mathematical induction.

THANKS

Is anything missing?

If you put n=10 or more or so, the inequality will not hold

**emakarov** · March 3rd 2010, 02:38 AM

I think the statement is correct. To prove it by induction, write the nth and (n+1)st sums: $s_n=1/(n+1)+\dots+1/(2n)$ ; $s_{n+1}=1/(n+2)+\dots+1/(2n+2)$ . Then express $s_{n+1}$ as $s_n+f(n)$ where $f(n)$ is some expression of n, i.e., $f(n)=s_{n+1}-s_n$ . If you show that $f(n)>0$ , you are done.

Note that $f(n)$ has three terms, and two of them are at least as big as half of the third.

**Archie Meade** · March 3rd 2010, 02:58 AM

Originally Posted by GamesonPlanes

Show 1/(n+1)+1/(n+2)+1/(n+3)+1/2n>3/5, for n>2. Use mathematical induction.

THANKS

Do you want to sum all terms from $\frac{1}{n+1}$ up to $\frac{1}{2n}$

not just the 4 terms quoted?

F(3)

$\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=\frac{9}{20}+\ frac{1}{6}=\frac{37}{60}>\frac{36}{60}$

True for n=3

F(k)

$\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+.....+\f rac{1}{2k}>\frac{3}{5}$ ?

F(k+1)

$\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac {1}{2k+1}+\frac{1}{2k+2}$

$=F(k)+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{1}{k+1}$

$=F(k)+\frac{1}{2k+1}+\frac{1}{2k+2}-\frac{2}{2k+2}$

$=F(k)+\frac{1}{2k+1}-\frac{1}{2k+2}$

A greater value in the denominator makes a smaller fraction, hence

$\frac{1}{2k+1}-\frac{1}{2k+2}>0$

Therefore if F(k) is true, F(k+1) certainly is

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