Computer Science Discrete Mathematics Proof problem

[bona0002]

Hey guys,

I'm taking Discrete Mathematics and am having a bit of trouble with one of my proofs. If any of you has any experience with that and could tell me where I'm going wrong, I'd appreciate it!

Ok, here it is:

Prove each statement in 8–23 by mathematical induction:

27. A sequence d1, d2, d3, . . . is defined by letting d1 = 2 and
dk = (d_(k−1)_)/k for all integers k ≥ 2. Show that for all integers
n ≥ 1, d_n_ = 2/n!

So here's what I got so far:

Proof: Let d_1_, d_2_, d_3_, ... be the sequence defined by specifying that d_1_ = 2 and d_k_ = (d_(k-1)_) / k for all integers k ≥ 2, and let the property P(n) be the equation:

d_n_ = 2 / n!

We will use mathematical induction to prove that for all integers n ≥ 1, P(n) is true.

Show that P(1) is true:

To establish P(1), we must show that d_1_ = 2 / 1!

But the left-hand side of d_1_ = 2 (by definition of d_1_, d_2, ...
and the right-hand side = 2 / 1! which is equal to 2/1 which is equal to 2.

Thus, the two sides of P(1) are equal, and hence P(1) is true.

Show that for all integers k ≥ 1, if P(k) is true then P(k+1) is also true.

Let k be any integer with k ≥ 1, and suppose P(k) is true. That is, suppose d_k_ = 2 / k!

We must show that P(k+1) is true. That is, we must show that d_(k+1)_ = 2 / (k+1)! which is equal to 2 / k! + 1! which is equal to 2 / k! + 1.

But the left-hand side of P(k+1) is d_(k+1)_ = d_k_ / (k+1) (By definition of sequence
which is, by the inductive step, equal to (2/k!) / ((k+1)/1) which is equal to 2/((k!)(k+1)).

Where do I go from there?

 

[Valkarie]

Thanks!</code>To show that P(k+1) is true, you need to show that the left-hand side is equal to the right-hand side. You have the left-hand side, so all you need to do is rearrange the right-hand side to get: 2/(k!(k+1)) = 2/((k+1)!).You can do this by multiplying both sides by (k+1)! and then cancelling out k! on both sides. Thus,2/(k!(k+1)) = 2*(k+1)!/(k!(k+1)!) = 2/(k+1)!which shows that P(k+1) is true.