Proving (1+1/n)^n=(n+1)^n/n! with Induction | Natural Numbers

[Boombaard]

Homework Statement

show with mathematical induction that prod(k=1->n) (1+1/n)^n=(n+1)^n/n!
n element Natural numbers

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?
thanks in advance :)ps. Happy New Year to those it applies t

 

[radou]

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!}
?

 

[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.

 

[Boombaard]

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!}
?

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

 

[radou]

yeah.. that'd be it

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.

 

[matt grime]

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

 

[Boombaard]

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

 

[radou]

@radou: i know it's supposed to be easy, but the answer still eludes me

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.

 

[Boombaard]

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.

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.

 

[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}

 

[Boombaard]

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}

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

 

[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.

 

[Boombaard]

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

thanks for your patience :)