# Help with induction problem (inequality)

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• October 11th 2010, 03:49 PM
kanoro
Help with induction problem (inequality)
Hi, problem is for any n >= 2, 1/n+1 + 1/n+2 + 1/n+3 + ... + 1/2n > 13/24 . I am having trouble putting into summation notation which is required first and then beginning. Help is appreciated, thanks
• October 11th 2010, 04:08 PM
Quote:

Originally Posted by kanoro
Hi, problem is for any n >= 2, 1/n+1 + 1/n+2 + 1/n+3 + ... + 1/2n > 13/24 . I am having trouble putting into summation notation which is required first and then beginning. Help is appreciated, thanks

$\displaystyle\sum_{m=1}^n\frac{1}{n+m}>\frac{13}{2 4},\;\;\;\;n\ \ge\ 2$

P(k)

$\displaystyle\sum_{m=1}^k\frac{1}{k+m}>\frac{13}{2 4}$

P(k+1)

$\displaystyle\sum_{m=1}^{k+1}\frac{1}{k+1+m}>\frac {13}{24}$

Try to prove that P(k+1) will be valid if P(k) is and test for the first n.
• October 11th 2010, 04:10 PM
raphw
Might not be a big thing, but do not forget to show that it is true for a specific n e.g. n = 2.
• October 11th 2010, 04:37 PM
kanoro
I understand it all, but when I am proving the initial case for n = 2, what value will the m be to make the inequality true?
• October 11th 2010, 04:41 PM
Quote:

Originally Posted by kanoro
I understand it all, but when I am proving the initial case for n = 2, what value will the m be to make the inequality true?

Your values of m go from 1 to n, so for n=2, you take the first 2 positive integers, m=1 and m=2.
• October 11th 2010, 04:43 PM
raphw
m is your index number for the sum. Thus m takes any value between 1 and n. The basic idea of Sigma ( $\sum$) is by example:

$\sum_{k=1}^{10} k = 1+2+3+4+5+6+7+8+9+10$

Just fill in values similarily-
• October 11th 2010, 04:51 PM
Quote:

Originally Posted by kanoro
I understand it all, but when I am proving the initial case for n = 2, what value will the m be to make the inequality true?

I forgot to answer the rest of this....

it isn't a matter of certain values of "m" making the inequality true.
You simply sum all the fractions (increasing m by 1 in each fraction) until m reaches n (not necessary when proving by induction).

For all n greater than or equal to 2, your sum ought to be greater than 13/24.
• October 11th 2010, 04:57 PM
kanoro
I'm working on solving the inequality but I am a little stuck.. I have the 1 / k + 1 + m > 13/24 and I understand the operations that can be applied to inequalities while maintaining to hold their values. Any clues on how to approach this?
• October 11th 2010, 05:18 PM
raphw
Note that:
$\displaystyle\sum_{m=1}^{k+1}\frac{1}{k+(1+m)} = \displaystyle\sum_{m=2}^{k+2}\frac{1}{k+m} = \displaystyle\sum_{m=1}^{k}\frac{1}{k+m} + \frac{1}{k+k+2} + \frac{1}{k+k+1}- \frac{1}{k+1}$

Now you should be able to solve.
• October 11th 2010, 05:47 PM
kanoro
Hmm, I am not seeing where this helps. Lost me there.
• October 11th 2010, 05:49 PM
Quote:

Originally Posted by kanoro
I'm working on solving the inequality but I am a little stuck.. I have the 1 / k + 1 + m > 13/24 and I understand the operations that can be applied to inequalities while maintaining to hold their values. Any clues on how to approach this?

You break down the P(k+1) to

$\displaystyle\sum_{m=1}^{k+1}\frac{1}{k+1+m}=\sum_ {m=1}^k\frac{1}{k+1+m}+\frac{1}{k+1+k+1}$

$=\displaystyle\sum_{m=1}^k\frac{1}{m+k}-\frac{1}{k+1}+\frac{1}{k+k+1}+\frac{1}{2k+2}$

which raphw has also shown and now you use P(k) to show that if P(k) is true,
then P(k+1) will also be true.

Add the three fractions on the right to show they sum to a value >0
• October 11th 2010, 06:08 PM
kanoro
So then I get summation as m=1 goes to k, 1/m+k - 8k+5 / ( 2k+2)(2k+1) . So how do I prove that they sum >0
• October 11th 2010, 06:37 PM
kanoro
In general still confused where to take this, textbook I am working out offers little help so any help to finish the question is much appreciated.
• October 11th 2010, 06:50 PM
raphw
The general idea of induction is as follows: (1) I show that something is true for a specific value of n, e.g. n=2. Then (2) I assume that this same statement is true for any n and (3) I show that it is then true for n+1. It follows that it must be true for any value since I can start by e.g. n=2 => true for n+1 = 3 => true for n+2 = 4 => ...

So we showed that it is true for n=2 already. Now we assume that it is true for $\displaystyle\sum_{m=1}^{k}\frac{1}{k+m}$

Thus we can solve
$\displaystyle\sum_{m=1}^{k+1}\frac{1}{k+(1+m)} = \displaystyle\sum_{m=2}^{k+2}\frac{1}{k+m} = \displaystyle\sum_{m=1}^{k}\frac{1}{k+m} + \frac{1}{k+k+2} + \frac{1}{k+k+1}- \frac{1}{k+1} > \frac{13}{24} + \frac{1}{k+k+2} + \frac{1}{k+k+1}- \frac{1}{k+1} > \frac{13}{24}$

Thus:
$\displaystyle\sum_{m=1}^{k+1}\frac{1}{k+(1+m)} > \frac{13}{24}$
and we completed our proof by induction since we can now construct any n as shown above.

This last step is true if and only if $\frac{1}{k+k+2} + \frac{1}{k+k+1}- \frac{1}{k+1} \ge 0$ This is all that is left to show to finish the proof. To do so, simply get a common denominator. If both numerator and denominator of the new, "merged" term are positive, you are done.
• October 11th 2010, 10:33 PM
Defunkt
Quote:

Originally Posted by raphw
The general idea of induction is as follows: (1) I show that something is true for a specific value of n, e.g. n=2. Then (2) I assume that this same statement is true for any n and (3) I show that it is then true for n+1. It follows that it must be true for any value since I can start by e.g. n=2 => true for n+1 = 3 => true for n+2 = 4 => ...

I know that this is semantics and you probably do understand that what you wrote is wrong - but what you're actually doing is proving that if the statement is true for some $n \ge 2$, then it is true for $n+1$. (Hi)
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