Proof of ##\sqrt[n]{n!}## ≤ ##\frac{n+1}{2}##

[chwala]

Yes, but you should prove that last inequality using a separate induction. To help you note that:
$$2(k+1)^{k+1} = 2(\frac{k+1}{k+2})^{k+1} (k+2)^{k+1}$$
If you can show that:
$$\forall k: \ \ 2(\frac{k+1}{k+2})^{k+1} \le 1$$
Then you are done.

the upper part of your post was a bit confusing...it would have been more clear if you had mentioned that we just divide both sides of my inequality in post $16$ by $(k+2)^{k+1}$, that makes sense to me...

 

[chwala]

What about:
$$2(k+1)^{k+1}≤ (k+2)^{k+1} \Leftrightarrow 2 \le (\frac{k+2}{k+1})^{k+1} = (1 + \frac 1 {k+1})^{k+1}$$

i am trying to follow your argument, ...following from my working,
$2(k+1)^{k+1}≤ (k+2)^{k+1}$...ends up to your working
$2$≤$(1 + \frac {1}{k+1})^{k+1}$ of course if $k>0$, then the inequality is satisfied, my only question is if step is really necessary...or are you of the opinion that even with this step that you've indicated, that the proof is not complete?

 

[PeroK]

$2$≤$(1 + \frac {1}{k+1})^{k+1}$ of course if $k>0$, then the inequality is satisfied

Why is this inequality satisfied?

 

[chwala]

Why is this inequality satisfied?

any value of $k>1$, will satisfy the inequality, for e.g
$k=1→ 2≤2.25$
$k=100→2≤2.7$
$k=1000000→2≤2.718$

 

[PeroK]

any value of $k>1$, will satisfy the inequality, for e.g
$k=1→ 2≤2.25$
$k=100→2≤2.7$
$k=1000000→2≤2.718$

That's not a proof. You could do that with the original inequality that you are taking all this effort to prove by induction!

 

[chwala]

That's not a proof. You could do that with the original inequality that you are taking all this effort to prove by induction!

perok i need your guidance ...i understand the steps, let me know exactly what i am supposed to do and from which step of my inequality.

 

[chwala]

i know that we now have $2≤ \frac {k+2}{k+1})^{k+1}≤ (1+ \frac{1}{k+1})^{k+1}$

 

[PeroK]

perok i need your guidance ...i understand the steps, let me know exactly what i am supposed to do and from which step of my inequality.

You need to prove that:
$$\forall k \ge 1: \ 2 \le (1 + \frac 1 {k+1})^{k+1}$$
And the hint was to use the binomial theorem. It might help to let $m = k+1$, then we need to show that:
$$\forall m \ge 2: \ 2 \le (1 + \frac 1 m)^m$$
Can you see how to apply the binomial theorem?

 

[chwala]

You need to prove that:
$$\forall k \ge 1: \ 2 \le (1 + \frac 1 {k+1})^{k+1}$$
And the hint was to use the binomial theorem. It might help to let $m = k+1$, then we need to show that:
$$\forall m \ge 2: \ 2 \le (1 + \frac 1 m)^m$$
Can you see how to apply the binomial theorem?

ok let me check this out...

 

[chwala]

ok let me check this out...

can i say $2^{1/m}≤(1 + \frac{1}{m})$

 

[chwala]

let me post another attempt, a minute...

 

[PeroK]

can i say $2^{1/m}≤(1 + \frac{1}{m})$

No. To be honest, all I'm doing here is giving you the solution step by step. The binomial theorem applied here gives:
$$(1 + \frac 1 m )^m = 1 + m(\frac 1 m) + \binom m 2 (\frac 1 m)^2 + \dots + (\frac 1 m)^m = 2 + \binom m 2 (\frac 1 m)^2 + \dots + (\frac 1 m)^m > 2 \ \ (m \ge 2)$$
The result is that I've done the whole solution now.

 

[chwala]

No. To be honest, all I'm doing here is giving you the solution step by step. The binomial theorem applied here gives:
$$(1 + \frac 1 m )^m = 1 + m(\frac 1 m) + \binom m 2 (\frac 1 m)^2 + \dots + (\frac 1 m)^m = 2 + \binom m 2 (\frac 1 m)^2 + \dots + (\frac 1 m)^m > 2 \ \ (m \ge 2)$$
The result is that I've done the whole solution now.

i was just about to type the same...lol

 

[chwala]

i had just researched on this...thanks perok...

 

[chwala]

He still needs to justify that last step.

he indicates that no justification is needed, you might need to respond in kind...we are all learning from your immense knowledge...

 

[haruspex]

he indicates that no justification is needed

Which strongly suggests he has been unable to do so.
Note that it is not true if we change the 2 in each denominator to e.

 

[chwala]

haruspex i am checking that out, on the side i find maths induction really really interesting. I am dedicating two weeks to study it. I would appreciate if i can get more easy to understand pdf. notes on mathematical induction for beginners...

 

[pasmith]

Homework Statement:: Using mathematical induction show that
$\sqrt[n]n!$ ≤$\frac{n+1}{2}$ where $n$ lies in $z^+$
Relevant Equations:: maths induction

$\sqrt[n]n!$ ≤$\frac{n+1}{2}$​

$\frac {1}{n}$ log [n!]= log$\frac{n+1}{2}$​

You could prove "by induction" that the AM-GM inequality applies to $\{1, \dots, n\}$, from which the result follows immediately...

 

[haruspex]

haruspex i am checking that out, on the side i find maths induction really really interesting. I am dedicating two weeks to study it. I would appreciate if i can get more easy to understand pdf. notes on mathematical induction for beginners...

As you know, one has to have an inductive hypothesis and a starting point.
In the induction problems given to students, these are both usually given.
In research, the hard part with induction can be picking the right hypothesis. You may set out to prove some property P, but to get the inductive step to work you might need to choose a stronger hypothesis.

 

[chwala]

You could prove "by induction" that the AM-GM inequality applies to $\{1, \dots, n\}$, from which the result follows immediately...

show me how...now that we already have a solution