# Yet another induction proof

1. Dec 31, 2006

### Boombaard

1. The problem statement, all variables and given/known data
show with mathematical induction that prod(k=1->n) (1+1/n)^n=(n+1)^n/n!
n element Natural numbers

3. The attempt at a solution

works for n=1,
$$\prod_{k=1}^{n+1}\left(1+\frac{1}{k+1}\right)^{k+1}$$ should become $$\frac{(n+2)^{n+1}}{(n+1)!}$$
$$= (1+(n+1)^{-1})*((1+(n+1)^{-1})^n)*\frac{(n+1)^n}{n!}$$
$$=\frac{(n+1)^n}{n!})*(1+(n+1)^{-1})(1+(n+1)^{-n})$$
$$=\frac{(n+1)^n}{n!}*(1+(n+1)^n+(n+1)^{-1}+(n+1)^{-n+1}$$
$$=\frac{(n+1)^n}{n!}+\frac{1}{n!}+\frac{(n+1)^n}{(n+1)!}+\frac{(n+1)^n}{(n+1)^{n+1})n!}$$
$$=\frac{(n+1)^{n+1}+(n+1)+(n+1)^n+1)}{(n+1)!}$$

only i can't find any way to get to (n+2)^(n+1)/(n+1)! from there :(
can anyone help?

ps. Happy New Year to those it applies t

Last edited: Jan 1, 2007
2. Dec 31, 2006

I might be missing something, but did you perhaps mean
$$\prod_{k=1}^n \left(1+\frac{1}{k}\right)^k = \frac{(n+1)^n}{n!}$$
?

3. Dec 31, 2006

### matt grime

I think you'd better take time out to learn latex if you're going to post things like that.

Anyway, you'd better get your n's and k's sorted out. I can't see any immediate interpretation of what yo'uve written that might be true.

4. Dec 31, 2006

### Boombaard

yeah.. that'd be it.. i tried editing the original a bit to make it more clear what exactly i've tried.. hope i didn't include any errors

Last edited: Dec 31, 2006
5. Dec 31, 2006

Ok, as you have seen, it holds for n = 1. Next, assume it holds for some n = j. You have to proove it holds for n = j + 1 in order to conclude it holds for every natural number. Now, write down what it looks like when n = j and when n = j + 1. The rest is easy.

6. Dec 31, 2006

### matt grime

It's easier to do without induction (just simplify the product), if you're allowed to. Doesn't sound like it.

7. Dec 31, 2006

### Boombaard

@radou: i know it's supposed to be easy, but the answer still eludes me
@matt: i'm not :)

8. Dec 31, 2006

Well, you have
$$\prod_{k=1}^{j+1}\left(1+\frac{1}{k}\right)^k = \prod_{k=1}^j \left(1+\frac{1}{k}\right)^k \cdot \left(1+\frac{1}{j+1}\right)^{j+1}$$.
That should illustrate the point, I hope.

9. Jan 1, 2007

### Boombaard

well, sure.. but, in my (granted, rather limited) experience with these things you then substitute the righthand side of part one with the product bit, and multiply that out, which should yield the n+1 RH part of the equation.. which is what i believe i tried to do in my try as shown above.. which is where i get stuck.

10. Jan 1, 2007

### matt grime

Then try again. That is all you need to do: mutliply out

$$\frac{(j+1)^{j}}{j!} \left(1 +\frac{1}{j+1}\right)^{j+1}$$

Last edited: Jan 1, 2007
11. Jan 1, 2007

### Boombaard

yes, but can you tell me if something has gone wrong in what i did in the OP? because it doesn't seem to add up to anything like the wanted result :S

Last edited: Jan 1, 2007
12. Jan 1, 2007

### matt grime

I cannot even decipher the original post - there is too much going on and things seem to vanish between lines, then reappear, like the factorial.

What is another way to write 1+ 1/(j+1)? It really is a one line simplification of the expression in my last post. If your original line of attack doesn't work try a different one.

13. Jan 1, 2007

### Boombaard

ah, yes.. i seem to have become somewhat obtuse after two years of not doing any maths :(