# Proof by induction, ##(n!)^{2} \le (2n)!##.

## Homework Statement

I need to prove by induction that ##(n!)^{2} \le (2n)!##. I'm pretty sure about my preliminary work, but I just need some suggestions for the end.

## Homework Equations

It is well known from a theorem that if ##a \le b## and ##c \ge 0##, then ##ca \le cb##.

## The Attempt at a Solution

ATTEMPT AT PROOF: For ##n=1##, we can see that ##(1!)^{2} \le (2\cdot 1)!## holds. Now, assume that ##(k!)^{2} \le (2k)!## holds for ##k \in N##. Then it must be shown that ##n=k+1## holds. For that,

##[(k+1)!]^{2} = (k+1)^{2}k!^{2}## and by assumption ##(k!)^{2} \le (2k)!## implies with the theorem above that

##(k+1)^{2}k!^{2} \le (k+1)^{2}(2k)!##

And this is the point at which I don't know how to proceed. I need to conclude that this is in fact less than ##[2(k+1)]!##, but I am unsure how to do that. I would appreciate any direction.

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

I need to prove by induction that ##(n!)^{2} \le (2n)!##. I'm pretty sure about my preliminary work, but I just need some suggestions for the end.

## Homework Equations

It is well known from a theorem that if ##a \le b## and ##c \ge 0##, then ##ca \le cb##.

## The Attempt at a Solution

ATTEMPT AT PROOF: For ##n=1##, we can see that ##(1!)^{2} \le (2\cdot 1)!## holds. Now, assume that ##(k!)^{2} \le (2k)!## holds for ##k \in N##. Then it must be shown that ##n=k+1## holds. For that,

##[(k+1)!]^{2} = (k+1)^{2}k!^{2}## and by assumption ##(k!)^{2} \le (2k)!## implies with the theorem above that

##(k+1)^{2}k!^{2} \le (k+1)^{2}(2k)!##

And this is the point at which I don't know how to proceed. I need to conclude that this is in fact less than ##[2(k+1)]!##, but I am unsure how to do that. I would appreciate any direction.
Are you absolutely required to use induction? I ask, because a direct proof (avoiding induction) is almost trivial.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
What is ##(2k)!/(2(k+1))!## ?

What is ##(2k)!/(2(k+1))!## ?

I assume you mean to use ##(2(k+1))!/(2(k+1))!## in which case the expression ##\frac{(2k)!}{(2(k+1))!}## is equivalent to ##\frac{1}{2(k+1)(2k+1)}##, and so this reduces my inequality to

##(k+1)^{2}(k!)^{2} \le \frac{k+1}{2(2k+1)}\cdot (2(k+1))! \le (2(k+1))!## since ##\frac{k+1}{2(2k+1)} < 1## for any ##k\in N##. Is that what you had in mind?

Are you absolutely required to use induction? I ask, because a direct proof (avoiding induction) is almost trivial.
I would prefer to use induction simply for the sake of practice. It is not required.

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Is that what you had in mind?
Yes.

PeroK
Homework Helper
Gold Member
2020 Award
##(k+1)^{2}k!^{2} \le (k+1)^{2}(2k)!##

And this is the point at which I don't know how to proceed. I need to conclude that this is in fact less than ##[2(k+1)]!##, but I am unsure how to do that. I would appreciate any direction.

If you expand the factorials, I think you'll see it fairly obvious.

• mikeyBoy83
If you expand the factorials, I think you'll see it fairly obvious.

Yes. You are right, thank you.