Question

\[\text { A sequence  } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and }  x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers } k, k \geq 2 . \text{ Show that }  x_n = \frac{2}{n!} \text{ for all } n \in N .\]

Answer 1

\[\text{ Given: A sequence } x_1 , x_2 , x_3 , . . . \text{ is defined by letting } x_1 = 2 \text{ and } x_k = \frac{x_{k - 1}}{k} \text{ for all natural numbers }  k, k \geq 2 . \]
\[\text{ Let }  P\left( n \right): x_n = \frac{2}{n!} \text{ for all } n \in N . \]
\[\text{ Step I: For } n = 1, \]
\[P\left( 1 \right): x_1 = \frac{2}{1!} = 2\]
\[\text{ So, it is true for n }  = 1 . \]
\[\text{ Step II: For n } = k, \]
\[\text{ Let } P\left( k \right): x_k = \frac{2}{k!} \text{ be true for some }  k \in N . \]
\[\text{ Step III: For n }  = k + 1, \]
\[P\left( k + 1 \right): \]
\[ x_{k + 1} = \frac{x_{k + 1 - 1}}{k}\]
\[ = \frac{x_k}{k}\]
\[ = \frac{2}{k \times k!} \left(\text {  Using step }  II \right)\]
\[ = \frac{2}{\left( k + 1 \right)!}\]
\[\text{ So, it is also true for n }  = k + 1 . \]
\[\text{ Hence,}  x_n = \frac{2}{n!} \text{ for all } n \in N .\]

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