sequence and product

Question

Show that $$\begin{cases} a_0=1 & \\ &\\ \forall n\in \mathbb{N}^*\quad a_{n+1}=\dfrac{a_n}{n(n+1)} & \end{cases} \underset{\implies}{\mbox{} } \fbox{$\forall n\in \mathbb{N}^*\quad a_{n}=\dfrac{a_1}{n!(n-1)!}$}$$

it's easy to show it by induction but i would ike to try the following way Indeed,

$$ \begin{aligned} \prod_{k=1}^{n-1} \frac{a_{n+1}}{a_n}&=\prod_{k=1}^{n-1}\dfrac{1}{k(k+1)}\\ \dfrac{a_{n}}{a_1}&=\prod_{k=1}^{n-1}\left( \dfrac{1}{k}-\dfrac{1}{k+1} \right)\\ & = \prod_{k=1}^{n-1}\frac{1}{k} - \prod_{k=1}^{n-1}\frac{1}{k+1} \end{aligned}$$

Answer 1

Your argument is fine up until the last line. You can't split up products of differences in this way. This would only work if you were dealing with sums instead of products.

Note that the product $\prod \limits_{i = 1}^2 (x_i - y_i)$ is equal to $(x_1 - y_1)(x_2 - y_2)$, while $\prod \limits_{i = 1}^2 x_i - \prod \limits_{i = 1}^2 y_i$ equals $x_1x_2 - y_1y_2$. These two terms are clearly different.

But products have the nice propert that they are commutative, i.e. you can complete your proof in the following way: $$\prod \limits_{k =1}^{n - 1} \frac{1}{k(k + 1)} = \left(\prod \limits_{k =1}^{n - 1} \frac{1}{k}\right)\left( \prod \limits_{k =1}^{n - 1} \frac{1}{k + 1}\right) = \left(\prod \limits_{k = 1}^{n - 1} k\right)^{-1} \left(\prod \limits_{k = 2}^n k\right)^{-1} = \frac{1}{n!(n - 1)!}.$$

Answer 2

it's easy to show it by induction but i would ike to try the following way

Note that "telescoping" like you do in $\prod_{k=1}^{n-1} \frac{a_{n+1}}{a_n} = \frac{a_{n}}{a_1}$ is in fact a disguised form of induction.

$$ \begin{aligned} \dfrac{a_{n}}{a_1}&=\prod_{k=1}^{n-1}\left( \dfrac{1}{k}-\dfrac{1}{k+1} \right) = \prod_{k=1}^{n-1}\frac{1}{k} - \prod_{k=1}^{n-1}\frac{1}{k+1} \end{aligned} $$

The second equality is wrong, since $\prod(a_k-b_k) \ne \prod a_k - \prod b_k$. What does follow, however, is:

$$ \begin{aligned} \dfrac{a_{n}}{a_1} = \prod_{k=1}^{n-1} \frac{a_{n+1}}{a_n} = \prod_{k=1}^{n-1}\dfrac{1}{k(k+1)} = \prod_{k=1}^{n-1}\dfrac{1}{k} \prod_{k=1}^{n-1}\dfrac{1}{k+1} = \dfrac{1}{\prod_{k=1}^{n-1} k} \;\dfrac{1}{\prod_{k=1}^{n-1} (k+1)} = \dfrac{1}{(n-1)!\,n!} \end{aligned} $$