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

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- Oct 11th 2010, 03:49 PMkanoroHelp 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

- Oct 11th 2010, 04:08 PMArchie Meade
$\displaystyle \displaystyle\sum_{m=1}^n\frac{1}{n+m}>\frac{13}{2 4},\;\;\;\;n\ \ge\ 2$

**P(k)**

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

**P(k+1)**

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

Try to prove that P(k+1) will be valid if P(k) is and test for the first n. - Oct 11th 2010, 04:10 PMraphw
Might not be a big thing, but do not forget to show that it is true for a specific n e.g. n = 2.

- Oct 11th 2010, 04:37 PMkanoro
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?

- Oct 11th 2010, 04:41 PMArchie Meade
- Oct 11th 2010, 04:43 PMraphw
m is your index number for the sum. Thus m takes any value between 1 and n. The basic idea of Sigma ($\displaystyle \sum$) is by example:

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

Just fill in values similarily- - Oct 11th 2010, 04:51 PMArchie Meade
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. - Oct 11th 2010, 04:57 PMkanoro
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?

- Oct 11th 2010, 05:18 PMraphw
Note that:

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

Now you should be able to solve. - Oct 11th 2010, 05:47 PMkanoro
Hmm, I am not seeing where this helps. Lost me there.

- Oct 11th 2010, 05:49 PMArchie Meade
You break down the P(k+1) to

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

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

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 - Oct 11th 2010, 06:08 PMkanoro
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

- Oct 11th 2010, 06:37 PMkanoro
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.

- Oct 11th 2010, 06:50 PMraphw
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 $\displaystyle \displaystyle\sum_{m=1}^{k}\frac{1}{k+m}$

Thus we can solve

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

Thus:

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

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 $\displaystyle \frac{1}{k+k+2} + \frac{1}{k+k+1}- \frac{1}{k+1} \ge 0$ 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. - Oct 11th 2010, 10:33 PMDefunkt