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