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!}$

  • $\begingroup$ Your initial statement is $P(2)$ not $P(1)$ $\endgroup$ – user10444 Apr 7 '14 at 16:56

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


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