# Thread: Problem 44

1. ## Problem 44

1)Let $n\geq 2$ prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.

Here is a problem for the younger kids, give them a chance, please.

2)Let two trains be located 100 miles from eachother. The trains travel to eachother at 30 miles per hour. At this moment a bird flys of one train and travels to the other at the speed of 60 miles per hour, the moment it reaches the train it turns around and goes back. It does it back and forth until the trains crash and annihilate the bird. Find the total distance the bird traveled.

2. Originally Posted by ThePerfectHacker
2)Let two trains be located 100 miles from eachother. The trains travel to eachother at 30 miles per hour. At this moment a bird flys of one train and travels to the other at the speed of 60 miles per hour, the moment it reaches the train it turns around and goes back. It does it back and forth until the trains crash and annihilate the bird. Find the total distance the bird traveled.
Dist. = speed x time

For one train to complete the journey:

$100 = \frac{30}{60 \ mins} \times t$

$t = 200$ mins

But now we have 2 trains, going at the same speed, so they will meet at 50 miles. So for the trains to meet each other will take 100 mins.

-----

Now for the bird.

The bird has 100 minutes to go as far as he can before his imminent and brutal death... (Oh the calamity)

$d = \frac{60}{60 \ mins} \times 100 \ mins$

$d = 100 \ miles$

EDIT: This looks like such a small distance. Something doesn't seem right.

EDIT2: Messed up with the time, let me correct it.

EDIT3: There we go, 100 miles

3. Originally Posted by janvdl
Dist. = speed x time

For one train to complete the journey:

$100 = \frac{30}{60 \ mins} \times t$

$t = 200$ mins

But now we have 2 trains, going at the same speed, so they will meet at 50 miles. So for the trains to meet each other will take 100 mins.

-----

Now for the bird.

The bird has 100 minutes to go as far as he can before his imminent and brutal death... (Oh the calamity)

$d = \frac{60}{60 \ mins} \times 100 \ mins$

$d = 100 \ miles$

EDIT: This looks like such a small distance. Something doesn't seem right.

EDIT2: Messed up with the time, let me correct it.

EDIT3: There we go, 100 miles
nice solution. another way to think about it:

two trains heading at each other at 30 mph each could be thought of as one train standing still and the other train heading toward it at 60 mph. the trains start at 100 miles apart. the bird flies off the train standing still and flies towards the train heading its way. the train and the bird meet halfway, that is, when each travel 50 miles, since they are moving at the same speed. once the bird touches the train and flies the other way, it does not pass the moving train, but stays exactly in front of it as it crashes into the standing train (since they are moving at the same speed). thus the bird flies 50 miles back to the standing train where it is crushed. thus the total distance the bird flies is 100 miles. this is a "logical solution," no knowledge of the formula for speed is necessary. it's pretty easy and quick to think through, so when you think for a few seconds and blurt out the answer, everyone can be impressed at your (apparent) calculating power. "How did he work that out so quickly?"

TPH, this question seems similar to one you've asked before. i think it was in another Problem of the Week thread.

4. Originally Posted by Jhevon
TPH, this question seems similar to one you've asked before. i think it was in another Problem of the Week thread.
Don't criticise it, at least it was the first problem of the week I was able to do. And I did it in such a physicsy manner. Topsquark's going to be so proud

5. Originally Posted by janvdl
Don't criticise it, at least it was the first problem of the week I was able to do. And I did it in such a physicsy manner. Topsquark's going to be so proud
I am!

-Dan

6. A lot of people, for some reason, do this problem by summing the infinite geometric series.

7. Originally Posted by ThePerfectHacker
A lot of people, for some reason, do this problem by summing the infinite geometric series.
To be honest, i thought about doing it that way, but i was unsure of how to go to work with it. How about showing that solution?

I was thinking the distance the bird travels will get smaller and smaller with time, and from there we could get the ratio. But that also seemed like a more tedious way to do it.

8. I like janvdl's answer. I would have never thought of that! I would have tried summing an infinite series, probably.

1)

Assume it is an integer:

1+ 1/2 + ... + 1/n = p/q
P and q have no common factors blah blah blah.

n + n/2 + ... + 1 = np/q
n(n-1) + n(n-1)/2 + ...+ 1 + (n-1) = n(n-1)p/q
n(n-1)(n-2) + n(n-1)(n-2)/2 + ...+ 1 + (n-2) + (n-1)(n-2) = n(n-1)(n-2)p/q

etc.

n! + (n-1)! = n!p/q (er- is this right?)
n + 1 = (n+1)p/q
i.f.f p=q, therefore assumption is wrong, blah blah blah?

Actually that's definitely wrong.

9. Originally Posted by ThePerfectHacker
1)Let $n\geq 2$ prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.
For n = 2, $1+\frac{1}{2}= \frac{3}{2}$ is not an integer.

Assume $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer for $2 \leq n \leq k$.

Then, take $n = k$, let

$1+\frac{1}{2}+...+\frac{1}{k}=\frac{k!+\frac{k!}{2 }+...+(k-1)!}{k!}=\frac{\sum_{x=1}^k \frac{k!}{x}}{k!}$

For this to not be an integer, we must have $k!\ \not|\ \sum_{x=1}^k \frac{k!}{x}$

So we must have

$\sum_{x=1}^k \frac{k!}{x}\neq ak!$ for some integer $a$. (and of course $k \geq 2$)

$k! \sum_{x=1}^k \frac{1}{x} \neq ak!$

$\sum_{x=1}^k \frac{1}{x} \neq a$

NOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!1

10. Originally Posted by DivideBy0
For n = 2, $1+\frac{1}{2}= \frac{3}{2}$ is not an integer.
...for some integer $a$....
$\sum_{x=1}^k \frac{1}{x} \neq a$

NOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!!!!!!!!!!!!!!!!!1
Dontcha just hate that >.<

11. Let $A_{n} = 1+\frac{1}{2}+...+\frac{1}{n}$.

Assume that, for some $n$, $A_{n}$ is an integer.

Let $P$ be the product of all the denominators except for the largest prime less than $n$. $P$ is clearly an integer.

Then $P \cdot A_{n} = P \cdot (1+\frac{1}{2}+...+\frac{1}{n})$.

On the LHS, $P \cdot A_{n}$ is an integer, based on how we defined P and our assumption.

On the RHS, after distributing, each term becomes an integer except for the term with the largest prime as the denominator. This clearly is not an integer, so the RHS is the sum of many integers plus one non-integer, and the RHS as a whole is not an integer.

Left with the integer LHS equaling a non-integer RHS, our assumption must be false, and $A_{n}$ must not be an integer.

$Q.E.D.$

12. Let me have a try:

problem:

1)Let $n\geq 2$ prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.

Solution:

If you can prove that $(\frac{1}{2} + ... + \frac{1}{n})$ is not an integer, then you can prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.

Why? if you add 1 to anything that's not an integer, you get a non-integer value.

look at this function:

$f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2}$

This rational function has asymptopes at the integers, which means that at every integer, the function approaches it, but never reaches it. This can be proven if we took the limit of the function as $x \rightarrow 1+, 2+, ....$ and so on. This means that the value of $f(n)$ never becomes an integer.

I conclude that $f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2} = (\frac{1}{2} + ... + \frac{1}{n})$

$1 + f(n)$ is not an integer, therefore $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer either.

Please tell me what you think.

13. Originally Posted by Skinner
Let me have a try:

problem:

1)Let $n\geq 2$ prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.

Solution:

If you can prove that $(\frac{1}{2} + ... + \frac{1}{n})$ is not an integer, then you can prove that $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer.

Why? if you add 1 to anything that's not an integer, you get a non-integer value.

look at this function:

$f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2}$

This rational function has asymptopes at the integers, which means that at every integer, the function approaches it, but never reaches it. This can be proven if we took the limit of the function as $x \rightarrow 1+, 2+, ....$ and so on. This means that the value of $f(n)$ never becomes an integer.

I conclude that $f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2} = (\frac{1}{2} + ... + \frac{1}{n})$

$1 + f(n)$ is not an integer, therefore $1+\frac{1}{2}+...+\frac{1}{n}$ is not an integer either.

Please tell me what you think.
You'll have to see what TPH thinks, but if your proof is right, i say Well Done! It's very easy to understand as well

14. Originally Posted by Skinner
$f(n) = \frac{1}{n} + \frac{1}{n-1} + \frac{1}{n-2} + ... + \frac{1}{2}$
How is this even a function? What is the value of $f(1/2)$? In order to talk about asymptotes you need to have a function defined at every single point (including $1/2$) except at the asymptote points.

15. oh. I messed up. Fail.

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