An identity of equivalent products related to the binomial theorem

**magus** · September 16th 2010, 08:08 PM

This was a step glossed over in a proof in the convergence of $a_n=(1+\frac{1}{n})^n$ that I can't seem to understand where certain terms are appearing from.

The identity is

$\dfrac{n(n-1)(n-2)\cdots(n-k+1)}{n^k}=\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) \cdots \left(1-\dfrac{k-1}{n}\right)$

I was able to get something similar by dividing $k$ $n$ s on the left hand side acros the 0th term $(n-0)$ to the $(k-1)$ st term but that still leaves the $k$ th and $(k+1)$ st term untouched and with a final term not in the form of the right hand side. It also leaves $(n-k)(n-k+1)$ in the left hand product.

Just to illustrate what I mean I will take the case of $k=4$

$\dfrac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{n*n*n*n}=\left(\dfrac{n}{n}\right)\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\left(1-\dfrac{3}{n}\right)(n-4)(n-5)$

Something tells me I'm making a wrong assumption somewhere or it's something dumb and obvious but I can't seem to get how this will end up in the form on the lhs.

Any help would be appreciated. Thank you for your time.

**tonio** · September 16th 2010, 08:38 PM

Originally Posted by magus

This was a step glossed over in a proof in the convergence of $a_n=(1+\frac{1}{n})^n$ that I can't seem to understand where certain terms are appearing from.

The identity is

$\dfrac{n(n-1)(n-2)\cdots(n-k+1)}{n^k}=\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right) \cdots \left(1-\dfrac{k-1}{n}\right)$

I was able to get something similar by dividing $k$ $n$ s on the left hand side acros the 0th term $(n-0)$ to the $(k-1)$ st term but that still leaves the $k$ th and $(k+1)$ st term untouched and with a final term not in the form of the right hand side. It also leaves $(n-k)(n-k+1)$ in the left hand product.

Just to illustrate what I mean I will take the case of $k=4$

$\dfrac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{n*n*n*n}=\left(\dfrac{n}{n}\right)\left(1-\dfrac{1}{n}\right)\left(1-\dfrac{2}{n}\right)\left(1-\dfrac{3}{n}\right)(n-4)(n-5)$

This is wrong: if $k=4$ then $n-k+1=n-4+1=n-3$ , and you reached up to 5...!

Tonio

Something tells me I'm making a wrong assumption somewhere or it's something dumb and obvious but I can't seem to get how this will end up in the form on the lhs.

Any help would be appreciated. Thank you for your time.

.

**magus** · September 16th 2010, 09:29 PM

But I thought that it was implied that the lhs product needed to go to the $(k+1)$ st term so I went up to 5

This is how I'm reading it.

$$\dfrac{\prod\limits^{k+1}_{i=0}(n-i)}{n^k}=\prod\limits^{k-1}_{i=1}\left( 1-\dfrac{i}{n}\right)$

Is that at least correct?

**Plato** · September 17th 2010, 03:01 AM

But $i=0,1,\cdots,k+1$ is $k+2$ values.

Whereas $i=1,2,\cdots,k-1$ is only $k-1$ values.

**tonio** · September 17th 2010, 05:38 AM

Originally Posted by magus

But I thought that it was implied that the lhs product needed to go to the $(k+1)$ st term so I went up to 5

I can't understand why or how you did understand that: the sequence $n,\,n-1,\,n-2,\ldots,\,n-k+1=n-(k-1)$

has $1+k-1=k$ summands, so with $k=4$ we get 4 summands...

I think you're confusing $n-(k+1)\,\,with\,\,n-k+1=n-(k-1)$ ...parentheses matter.

Tonio

This is how I'm reading it.

$$\dfrac{\prod\limits^{k+1}_{i=0}(n-i)}{n^k}=\prod\limits^{k-1}_{i=1}\left( 1-\dfrac{i}{n}\right)$

Is that at least correct?

.

**Prove It** · September 17th 2010, 07:48 AM

Note that with

$\frac{n(n-1)(n-2)\dots(n-k+1)}{n^k}$ has $k$ factors in the numerator.

So you can write it as...

$\frac{n(n-1)(n-2)\dots(n-k+1)}{n^k} = \left(\frac{n}{n}\right)\left(\frac{n - 1}{n}\right)\left(\frac{n - 2}{n}\right)\dots\left[\frac{n - (k - 1)}{n}\right]$

$= 1\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\dots\left(1 - \frac{k - 1}{n}\right)$ as required.

**magus** · September 17th 2010, 12:03 PM

Ok I'm just done with this. I can't see how to go from a $k+1$ to a $k-1$ on the other side. Screw this, I found a product formula: induction it is. This is what I get for trying to learn analysis on my own.

Thank you all for all your help and I know it's something obvious that I've gotten backwards somewhere but it just isn't clicking so I'm going to try induction.

**tonio** · September 18th 2010, 03:34 AM

Originally Posted by magus

Ok I'm just done with this. I can't see how to go from a $k+1$ to a $k-1$ on the other side. Screw this, I found a product formula: induction it is. This is what I get for trying to learn analysis on my own.

Thank you all for all your help and I know it's something obvious that I've gotten backwards somewhere but it just isn't clicking so I'm going to try induction.

You can try induction as much as you can try transcendental yoga: if you still don't understand why

there are only k factors in that product, instead of k+1, and that the last one is n-k+1 = n-(k-1) ,

then you won't be able to complete the induction because you're gonna choose the wrong values...Good luck, though!

Tonio

**magus** · September 19th 2010, 11:53 PM

Originally Posted by tonio

You can try induction as much as you can try transcendental yoga: if you still don't understand why

there are only k factors in that product, instead of k+1, and that the last one is n-k+1 = n-(k-1) ,

then you won't be able to complete the induction because you're gonna choose the wrong values...Good luck, though!

Tonio

I understand it now, but I did it with induction using the product notation anyway because it made more sense to me that way. Thank you for your help.

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