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Yet another induction proof

  • Thread starter Boombaard
  • Start date
10
0
1. Homework Statement
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,
[tex]\prod_{k=1}^{n+1}\left(1+\frac{1}{k+1}\right)^{k+1}[/tex] should become [tex]\frac{(n+2)^{n+1}}{(n+1)!}[/tex]
[tex]= (1+(n+1)^{-1})*((1+(n+1)^{-1})^n)*\frac{(n+1)^n}{n!}[/tex]
[tex]=\frac{(n+1)^n}{n!})*(1+(n+1)^{-1})(1+(n+1)^{-n})[/tex]
[tex]=\frac{(n+1)^n}{n!}*(1+(n+1)^n+(n+1)^{-1}+(n+1)^{-n+1}[/tex]
[tex]=\frac{(n+1)^n}{n!}+\frac{1}{n!}+\frac{(n+1)^n}{(n+1)!}+\frac{(n+1)^n}{(n+1)^{n+1})n!}[/tex]
[tex]=\frac{(n+1)^{n+1}+(n+1)+(n+1)^n+1)}{(n+1)!}[/tex]

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 to:)
 
Last edited:

radou

Homework Helper
3,105
6
I might be missing something, but did you perhaps mean
[tex]\prod_{k=1}^n \left(1+\frac{1}{k}\right)^k = \frac{(n+1)^n}{n!}[/tex]
?
 

matt grime

Science Advisor
Homework Helper
9,394
3
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.
 
10
0
I might be missing something, but did you perhaps mean
[tex]\prod_{k=1}^n \left(1+\frac{1}{k}\right)^k = \frac{(n+1)^n}{n!}[/tex]
?
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:

radou

Homework Helper
3,105
6
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

Science Advisor
Homework Helper
9,394
3
It's easier to do without induction (just simplify the product), if you're allowed to. Doesn't sound like it.
 
10
0
@radou: i know it's supposed to be easy, but the answer still eludes me
@matt: i'm not :)
 

radou

Homework Helper
3,105
6
@radou: i know it's supposed to be easy, but the answer still eludes me
Well, you have
[tex]\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}[/tex].
That should illustrate the point, I hope.
 
10
0
Well, you have
[tex]\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}[/tex].
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

Science Advisor
Homework Helper
9,394
3
Then try again. That is all you need to do: mutliply out

[tex] \frac{(j+1)^{j}}{j!} \left(1 +\frac{1}{j+1}\right)^{j+1}[/tex]
 
Last edited:
10
0
Then try again. That is all you need to do: mutliply out

[tex] \frac{(j+1)^{j}}{j!} \left(1 +\frac{1}{j+1}\right)^{j+1}[/tex]
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:

matt grime

Science Advisor
Homework Helper
9,394
3
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.
 
10
0
ah, yes.. i seem to have become somewhat obtuse after two years of not doing any maths :(

thanks for your patience :)
 

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