# explain this simplification (mathematical induction and sequences)

a sequence $a_1,a_2,a_3,...$ is defined by letting $a_1 = 2$ and $a_k = \frac{a_k-1}{k}$ for all integers $k\ge2$.

use mathematical induction to prove $a_n = \frac{2}{n!}$ for all integers $n\ge1$ is a general form of the sequence.

let $p(1)$ be the statement $a_1 = \frac{2}{1!} = 2$

now the left hand side of $p(1)$ is 2. the right hand side is also 2 so the statement holds.

now assume $p(k)$ holds for $k\ge1$ so assume that $a_k = \frac{2}{k!}$

the statement $p(k+1)$ states: $a_k+1 = \frac{2}{(k+1)!}$

the left hand side of $$p(k+1) = a_k+1$$ $$= \frac{a_k}{k+1}$$ $$= \frac{2}{k!(k+1)}$$ $$= \frac{2}{k!(k+1)}$$ $$= \frac{2}{(k+1)!}$$ which equals right hand side of $p(k+1)$

can someone please explain to me how $\frac{2}{k!(k+1)}$ becomes $\frac{2}{(k+1)!}$?

I thought that $\frac{2}{k!(k+1)}$ would become $\frac{2}{k^2!+k!}$

It's just the definition of the factorial: $$(k+1)!=1\cdot2\cdot...\cdot k\cdot (k+1)=k!(k+1)$$