I'm stuck on this homework problem. I must prove the statement using mathematical induction
Given: A sequence $d_1, d_2, d_3, ...$ is defined by letting $d_1 = 2$ and for all integers k $\ge$ 2. $$ d_k = \frac{d_{k-1}}{k} $$
Show that for all integers $n \ge 1$ , $$d_n = \frac{2}{n!}$$
Here's my work:
Proof (by mathematical induction). For the given statement, let the property $p(n)$ be the equation:
$$ d_n = \frac{2}{n!} $$
Show that $P(1)$ is true: The left hand side of $P(1)$ is $d_n$ , which equals $2$ by definition of the sequence. The right hand side is:
$$ \frac{2}{(1)!} =2 $$
Show for all integers $k \geq 1$, if $P(k)$ is true, then $p(k+1)$ is true. Let k be any integer with $k \geq 1$, and suppose $P(k)$ is true. That is, suppose: (This is the inductive hypothesis)
$$ d_{k} = \frac{2}{k!} $$
We must show that $P(K+1)$ is true. That is, we must show that:
$$ d_{k+1} = \frac{2}{(k+1)!} $$
(I thought I was good until here.)
But the left hand side of $P(k+1)$ is:
$$ d_{k+1} = \frac{d_k}{k+1} $$
By inductive hypothesis:
$$ d_{k+1} = \frac{(\frac{2}{2!})}{k+1} $$
$$ d_{k+1} = \frac{2}{2!}\frac{1}{k+1} $$
but that doesn't seem to equal what I needed to prove: $ d_n = \frac{2}{n!}$
The following is not true $$d_{k+1} = \frac{(\frac{2}{2!})}{k+1}$$ since $d_k=\frac{2}{k!}$ not $\frac{2}{2!}$, you actually have $$d_{k+1} = \frac{(\frac{2}{k!})}{k+1}=\frac{(\frac{2}{k!(k+1)})}{1}=\frac{2}{(k+1)!}$$