Solve the following
Question:

A sequence $x_{1}, x_{2}, x_{3}, \ldots$ is defined by letting $x_{1}=2$ and $x_{k}=\frac{x_{k-1}}{k}$ for all natural numbers $k, k \geq 2 .$ Show that $x_{n}=\frac{2}{n !}$ for all $n \in \mathbf{N}$.

Solution:

Given : A sequence $x_{1}, x_{2}, x_{3}, \ldots$ is defined by letting $x_{1}=2$ and $x_{k}=\frac{x_{k-1}}{k}$ for all natural numbers $k, k \geq 2$.

Let $\mathrm{P}(n): x_{n}=\frac{2}{n !}$ for all $n \in \mathbf{N}$.

Step I: For $n=1$,

$\mathrm{P}(1): x_{1}=\frac{2}{1 !}=2$

So, it is true for $n=1$.

Step II : For $n=k$,

Let $\mathrm{P}(k): x_{k}=\frac{2}{k !}$ be true for some $k \in \mathbf{N}$.

Step III : For $n=k+1$,

$\mathrm{P}(k+1)$ :

$x_{k+1}=\frac{x_{k+1-1}}{k}$

$=\frac{2}{k \times k !} \quad($ Using step II $)$

$=\frac{2}{(k+1) !}$

So, it is also true for $n=k+1$.

Hence, $x_{n}=\frac{2}{n !}$ for all $n \in \mathbf{N}$.

Disclaimer: It should be $k$ instead $n$ in the denominator of $x_{k}=\frac{x_{k-1}}{k}$. The same has been corrected above.