Proof by induction of infinite product

Question

I just started learning inductive proofs but I am stuck trying to proof that the following equation applies for all $n\in\mathbb{N}_0$. \begin{align} n!&=\prod_{i=1}^{\infty}{\left[\left(\frac{i+1}{i}\right)^n\frac{i}{n+i}\right]} \end{align}

I already showed the base case \begin{align} 0!=\prod_{i=1}^{\infty}{\left[\left(\frac{i+1}{i}\right)^0\frac{i}{i}\right]}=1 \end{align} and thus we can assume \begin{align} k!=\prod_{i=1}^{\infty}{\left[\left(\frac{i+1}{i}\right)^k\frac{i}{k+i}\right]} \end{align} But I am having issues with the inductive step $k\to k+1$. \begin{align} (k+1)!&=(k+1)k!\\ &=(k+1)\prod_{i=1}^{\infty}{\left[\left(\frac{i+1}{i}\right)^k\frac{i}{k+i}\right]}\\ &=\cdots \end{align} I do not know how to continue with this. Is it possible to move $(k+1)$ inside the product? Can anyone please help me with this problem. Thanks in advance!

Answer 1

First off, you cannot move $k+1$ inside the product; consider $2 \cdot \prod_{i=1}^{3} i=2 (1) (2) (3)= 12$. This is not equivalent to $\prod_{i=1}^{3} (2i) = (2) (4) (6) = 48$.

Also, the inductive step works a bit differently. In this case, we want to prove that $\prod_{i=1}^{\infty} (\frac{i+1}{i})^{k+1} \frac{i}{(k+1)+i} = (k+1) \prod_{i=1}^{\infty} (\frac{i+1}{i})^{k} \frac{i}{k+i}$. This is equivalent to showing that we can manipulate $(k+1) k!$(right side) into $(k+1)!$(left side. If we can do that, we have achieved the definition of the factorial.

We can start by splitting the product: we get

$\prod_{i=1}^{\infty} (\frac{i+1}{i})^{k+1} \prod_{i=1}^{\infty} \frac{i}{(k+1)+i} = (k+1) \prod_{i=1}^{\infty} (\frac{i+1}{i})^{k} \prod_{i=1}^{\infty} \frac{i}{k+i}$

Now, noting that the left and right hand sides are extremely similar, we consider dividing both sides by the products on the right.

$\frac{\prod_{i=1}^{\infty} (\frac{i+1}{i})^{k+1} \prod_{i=1}^{\infty} \frac{i}{(k+1)+i}}{\prod_{i=1}^{\infty} (\frac{i+1}{i})^{k} \prod_{i=1}^{\infty} \frac{i}{k+i}} = k+1$

While this looks ugly, we notice that we can combine the similar products. Consider $\frac{\prod_{i=1}^{3} i}{\prod_{i=1}^{3} 2i}=\frac{(1) (2) (3)}{(2) (4) (6)} = \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \prod_{i=1}^{3} \frac{i}{2i}$. Doing this, we have:

$\prod_{i=1}^{\infty} \frac{i+1}{i} \prod_{i=1}^{\infty} \frac{k+i}{k+1+i} = k+1$

$\prod_{i=1}^{\infty} \frac{(i+1) (k+i)}{i(k+1+i)} = k+1$

$\prod_{i=1}^{\infty} (1 + \frac{k}{i(k+i+1)}) = k+1$

To find the product on the left, we claim that the partial products $\prod_{i=1}^{m} (1 + \frac{k}{i(k+i+1)})$ are equal to $\frac{(k+1)(m+1)}{k+m+1}$. This is very easy to prove through induction and is left as an exercise. (EDIT: You need not use partial products, the product trivially telescopes as shown in another answer).

Now, we just need to take the limit as $m$ goes to infinity. This will give us

$\lim_{m \to \infty} \frac{(k+1)(m+1)}{k+m+1} =$

$(k+1) \lim_{m \to \infty} \frac{m+1}{k+m+1} =$

$(k+1) \cdot 1 =$

$k+1$

And hence the inductive step is true, and hence $n! =\prod_{k=1}^{\infty} (\frac{i+1}{i})^k \frac{i}{n+i}$ for all integer $n$.

Sidenote: I tried to find a partial product formula for the original product, but could not.

Answer 2

Look at the product

$\prod\limits_{i=1}^\infty \dfrac{(k+i)(i+1)}{i(k+1+i)}=\dfrac{2(k+1)}{(k+2)}\dfrac{3(k+2)}{2(k+3)}\dfrac{4(k+3)}{3(k+4)}\ldots=k+1$

Hence you have

$\prod\limits_{i=1}^\infty \left(\dfrac{i+1}{i}\right)^{k+1}\dfrac{i}{(k+1)+i}= \prod\limits_{i=1}^\infty \left(\dfrac{i+1}{i}\right)^{k} \dfrac{i+1}{i}\dfrac{i}{(k+1)+i}= \prod\limits_{i=1}^\infty \left(\dfrac{i+1}{i}\right)^{k} \dfrac{i}{k+i}\dfrac{k+i}{i}\dfrac{i+1}{i}\dfrac{i}{(k+1)+i}= \prod\limits_{i=1}^\infty \left(\dfrac{i+1}{i}\right)^{k} \dfrac{i}{k+i}\prod\limits_{i=1}^\infty\dfrac{k+i}{i}\dfrac{i+1}{i}\dfrac{i}{(k+1)+i} \overset{\text{Induction hypotheses}}=k!\prod\limits_{i=1}^\infty \dfrac{(k+i)(i+1)}{i(k+1+i)}=(k+1)!$

The nice thing about these induction proofs is, that you already know that the given equation is true. Hence if some unwanted terms remain, you already know what they have to be. In this case in particular it was clear that the remaining product $\prod\limits_{i=1}^\infty \dfrac{(k+i)(i+1)}{i(k+1+i)}$ must equal $k+1$.

Answer 3

Here we continue the induction step as started by OP.

We obtain \begin{align*} \color{blue}{(k+1)!}&=(k+1)k!\\ &=(k+1)\prod_{k=1}^{\infty}\left(\left(\frac{i+1}{i}\right)^k\frac{i}{k+i}\right)\tag{1}\\ &=(k+1)\lim_{K\to\infty}\prod_{k=1}^{K}\left(\left(\frac{i+1}{i}\right)^k\frac{i}{k+i}\right)\tag{2}\\ &=(k+1)\lim_{K\to\infty}\left(\prod_{k=1}^{K}\left(\left(\frac{i+1}{i}\right)^{k+1}\frac{i}{k+1+i}\right)\right.\\ &\qquad\qquad\qquad\qquad\left.\cdot\prod_{i=1}^K\left(\frac{i}{i+1}\frac{k+1+i}{k+i}\right)\right)\tag{3}\\ &=(k+1)\prod_{k=1}^{\infty}\left(\left(\frac{i+1}{i}\right)^{k+1}\frac{i}{k+1+i}\right)\\ &\qquad\qquad\qquad\qquad\cdot\lim_{K\to\infty}\left(\frac{K!}{(K+1)!}\,\frac{(k+1+K)!}{(k+1)!}\,\frac{k!}{(k+K)!}\right)\tag{4}\\ &=\prod_{k=1}^{\infty}\left(\left(\frac{i+1}{i}\right)^{k+1}\frac{i}{k+1+i}\right) \lim_{K\to\infty}\frac{k+1+K}{K+1}\tag{5}\\ &=\prod_{k=1}^{\infty}\left(\left(\frac{i+1}{i}\right)^{k+1}\frac{i}{k+1+i}\right)\lim_{K\to\infty}\left(1+\frac{k}{K+1}\right)\\ &\,\,\color{blue}{=\prod_{k=1}^{\infty}\left(\left(\frac{i+1}{i}\right)^{k+1}\frac{i}{k+1+i}\right)} \end{align*} and the induction step follows.

Comment:

• In (1) we use the induction hypothesis for $k!$.

• In (2) we use the definition of an infinite product as limit of a finite product.

• In (3) we multiply the product with the factors we need to perform the step $k\to k+1$ and we multiply with a second product with corresponding terms for compensation.

• In (4) we split the limit, which is admissible if we can show that both of them exist. We also cancel terms in the right-hand product, since using factorial notation we get \begin{align*} \prod_{i=1}^K\frac{i}{i+1}&=\frac{K!}{(K+1)!}=\frac{1}{K+1}\\ \prod_{i=1}^K\frac{k+1+i}{k+i}&=\prod_{i=1}^K(k+1+i)\prod_{i=1}^K\frac{1}{k+i}\\ &=\frac{(k+1+K)!}{(k+1)!}\frac{k!}{(k+K)!}\\ &=\frac{k+1+K}{k+1} \end{align*}

• In (5) we do some further cancellation. Note that the factor $k+1$ from the left-hand side of the expression is also cancelled here.