Prove that for $n \ge 2$ the follow inequality holds $\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}$.

Question

As already shown above I need to prove that $$\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}\qquad \forall n \ge 2.$$

What I've come up with is the following:

$\underline{n=2:}\qquad$ $$\frac {16}{3} \lt 6$$

$\underline{n=k:}\qquad $$$\frac {4^k}{k+1} \lt \frac{(2k)!}{(k!)^2} $$

$\underline{n+1}:$ $$\frac {4^{k+1}}{k+2} \lt \frac{(2(k+1))!}{((k+1)!)^2} $$ $$\frac{4^k \cdot 4}{k+2} \lt \frac {(2k+2)!}{(k!)^2\cdot(k+1)\cdot(k+1)}$$ $$\frac{4^k \cdot 4}{k+2} \lt \frac{(k+1)\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)\cdot(2k+2)}{(k!)^\require{enclose}\enclose{updiagonalstrike}2\cdot(k+1)\cdot(k+1)}$$ $$\frac{4^k \cdot 4}{k+2} \lt 2 \cdot\frac{\require{enclose}\enclose{updiagonalstrike}{(k+1)}\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}}{(k!)\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}\cdot\require{enclose}\enclose{updiagonalstrike}{(k+1)}}$$ $$\frac{4^k \cdot 4}{k+2} \lt 2 \cdot\frac{{(k+2)}\cdot \ldots \cdot (k + (k-1))\cdot{2k} \cdot (2k+1)}{(k!)}$$

From here on I'm stuck. Can somebody help me please?

In the 6th line of equations, you cancelled off the $k!$ in $\frac{(2k)!}{({k!})^2}$. You cannot do that because $(2k)!$ does not equal $2(k!)$ — Harry Alli, Sep 20 '17 at 13:37

score 2 · Accepted Answer · 2017-09-20T20:01:32.170

2

For $k+1$ it is equivalent to

$$ \frac{4^k}{k+1}<\frac{(2k+1)(k+2)(2k)!}{2(k+1)^2(k!)^2}$$

It is enough to prove that

$$\frac{(2k+1)(k+2)}{2(k+1)^2}>1$$

Notice that the above fraction is monotonically decreasing and its limit at $k\to\infty$ is $1$, therefore it is always greater than $1$.

And also, you have a mistake, as Harry noticed, $(2k)!\neq2k!$.

Edit

I have no idea what is Jaroslaw talking about. There is step-by-step solution. Suppose it is true for $k$. Then it is true for $k+1$ if and only if

$$\frac{4^{k+1}}{k+2}<\frac{(2k+2)!}{\left((k+1)!\right)^2}\\\frac{4^k}{k+1}\cdot\frac{4(k+1)}{k+2}<\frac{(2k+2)(2k+1)(2k)!}{(k+1)^2(k!)^2}\\\frac{4^k}{k+1}\cdot\frac{4(k+1)}{k+2}<\frac{(2k)!}{(k!)^2}\cdot\frac{(2k+2)(2k+1)}{(k+1)^2}$$

It is enough to prove that:

$$\frac{4(k+1)}{k+2}<\frac{(2k+2)(2k+1)}{(k+1)^2}$$

It is equivalent to

$$\frac{(2k+2)(2k+1)}{(k+1)^2}>\frac{4(k+1)}{k+2}\\\frac{2(k+1)(2k+1)(k+2)}{4(k+1)(k+1)^2}>1\\\frac{(2k+1)(k+2)}{2(k+1)^2}>1$$

edited Sep 20 '17 at 20:01

answered Sep 20 '17 at 13:35

My apologies for that. I'll edit that – Anonymous196 Sep 20 '17 at 13:50
Your inequality, that You say is equivalent, actually isn't equivalent. On the left side of equivalent inequality there should be $\frac{4^{k+1}}{k+2}$ and on the right side $\frac{(2k+1)(k+\color{red}{1})(2k)!}{2(k+1)^2(k!)^2}$ – Jaroslaw Matlak Sep 20 '17 at 15:54
@JaroslawMatlak. Actually the inequalities ARE equivalent, because we have that $k\ge3$. There is logical equivalency between original inequality for $k+1$ and my inequality. Why downvote? Do you want a proof or what? Please reconsider your vote. Thanks. – Sep 20 '17 at 18:22
For the left side - I really don't know how inequality $a<b$ could be equivalent to $c<b$, especially when $a<c$. For the right side - obviously your version is an one way implication. Can you proove the other way implication? And also it doesn't give us the expected right side. – Jaroslaw Matlak Sep 20 '17 at 18:43
@JaroslawMatlak. I don't understand what do you mean. Please take a look at my answer again, I added an edit. I don't see what is incorrect in my implications. And, please reconsider your vote. Thanks. – Sep 20 '17 at 20:02
@ThePirateBay Now your reasoning is clear for me. Thanks. – Jaroslaw Matlak Sep 20 '17 at 20:34

score 1 · Answer 2 · edited Sep 20 '17 at 15:04

1

Note that $$\frac{\frac{4^{n+1}}{n+2}}{\frac{4^n}{n+1}}=4\frac{n+1}{n+2}$$ and that $$\frac{\frac{(2(n+1))!}{(n+1)!^2}}{\frac{(2n)!}{n!^2}}=\frac{(2n+2)!}{(2n)!}\cdot\frac{n!^2}{(n+1)!^2}=\frac{(2n+2)(2n+1)}{(n+1)^2}=2\frac{2n+1}{n+1}. $$So, when is it true that$$4\frac{n+1}{n+2}\leqslant2\frac{2n+1}{n+1}?$$ Well, the previous inequaliy is equivalent to $2(n+1)^2\leqslant(2n+1)(n+2)=2n^2+5n+2$, which clearly holds always. So, since you proved that $\frac{4^2}{2+1}<\frac{(2\times2)!}{2!^2}$, the inequality that you want to prove is true whenever $n\geqslant2$.

edited Sep 20 '17 at 15:04

Anonymous196

1,385

answered Sep 20 '17 at 13:36

José Carlos Santos

427,504

Is this way of proving considered as an induction proof or a heuristics proof? Because you used the $P(n+1)/ P(n)$-technique with P is the statement above. – Anonymous196 Sep 20 '17 at 14:08
@AnonymousI What I provided is an almost complete proof by induction. Let $a_n=\frac{4^n}{n+1}$ and let $b_n=\frac{(2n)!}{n!^2}$. In order to complete the proof from what I did, the only step missing is to observe that, if $a_n<b_n$, then $a_{n+1}=a_n\times\frac{a_{n+1}}{a_n}<b_n\times\frac{b_{n+1}}{b_n}=b_{n+1}$. – José Carlos Santos Sep 20 '17 at 14:12
I'm just used to the conventional way of using induction without any tricks. I need a lot more of exercises to get the hang of induction. I'll try some math olympiads questions. – Anonymous196 Sep 20 '17 at 14:17
@AnonymousI This is not a trick! It is the standard way of proving statements of the type $(\forall n\in\mathbb{N}):a_n<b_n$ (or $a_n\leqslant b_n$) when the $a_n$'s and the $b_n$'s are defined in a multiplicative way. – José Carlos Santos Sep 20 '17 at 14:19
Oké, I just need more exercises to work on. My apologies for the previous remark. – Anonymous196 Sep 20 '17 at 14:38
Where does the last 2 between** come from in $2(n+1)^2\leqslant2(2n+1)(n+2)=4n^2+12n+4$ – Anonymous196 Sep 20 '17 at 14:52
@AnonymousI It shouldn't be there. My mistake. I've edited my answer. Please tell me if all is well now. – José Carlos Santos Sep 20 '17 at 14:56
I'd think that the 4 is divided by the 2 in $4\frac{n+1}{n+2}\leqslant2\frac{2n+1}{n+1}$ and that's it – Anonymous196 Sep 20 '17 at 14:56
Yep everything is ok. – Anonymous196 Sep 20 '17 at 14:57
@AnonymousI That's what I did. And then I multiplied both sides by $(n+1)(n+2)$. – José Carlos Santos Sep 20 '17 at 14:57
Thank you very much for your attention and patience – Anonymous196 Sep 20 '17 at 14:57
I've another question: If you say that P(n+1)/ P(n) is that similar to this. Or is it completely false. – Anonymous196 Oct 19 '17 at 09:41
@AnonymousI I have no idea about what you mean by this. – José Carlos Santos Oct 19 '17 at 09:44
Your approach is just not directly similar to the basic induction proofs for me. I just have a hard time with writing it down formally. – Anonymous196 Oct 19 '17 at 09:45
I think I now see the missing link which is the induction hypothesis for $a_n$ and $b_n$ in $a_{n+1} =a_n \times \frac {a_{n+1}}{a_n} \lt b_n \times \frac {b_{n+1}}{b_n}=b_{n+1}$ without those in the fractions. – Anonymous196 Oct 19 '17 at 09:48
@AnonymousI Then re-write my approach in a way which is similar to the basic induction proofs that you are aware of. Besides, why are you even interested in my method? You marked another answer as the accepted one. – José Carlos Santos Oct 19 '17 at 09:48
I think I see it now. I think I have defined asking stupid questions on a whole new way. I can't stress enough how much I appreciate your help. – Anonymous196 Oct 19 '17 at 09:49
Do you have any suggestions to good sites or good books on how to solve inequalities? – Anonymous196 Oct 19 '17 at 10:02
@AnonymousI No. I am not strong on inequalities. – José Carlos Santos Oct 19 '17 at 10:04
Oh, ok. Have a good day, sir. – Anonymous196 Oct 19 '17 at 10:05

Jaroslaw Matlak · Answer 3 · 2017-09-20T16:05:22.910

1

As your first steps of mathematical induction are right, I'll just perform a proof of inductive thesis for $n=k+1$.

First notice, that for $k>1$ $$\frac{4(k+1)}{(k+2)} < \frac{(2k+2)(2k+1)}{(k+1)^2}$$ proof: $$(k-1)k(k+1)^2(k+2)>0\\ \frac{k(k-1)}{(k+1)^2(k+2)}>0\\ \frac{4k^2-4k}{(k+1)^2(k+2)}>0\\ \frac{-4k^2+4k}{(k+1)^2(k+2)}<0\\ \frac{4(k^3+3k^2+3k+1)-(4k^3+16k^2+8k+4)}{(k+1)^2(k+2)}<0\\ \frac{4(k+1)^3}{(k+1)^2(k+2)}<\frac{4k^3+16k^2+8k+4}{(k+1)^2(k+2)}\\ \frac{4(k+1)^2}{(k+1)^2(k+2)}<\frac{(2k+2)(2k+1)(k+2)}{(k+1)^2(k+2)}\\ \frac{4(k+1)}{(k+2)}<\frac{(2k+2)(2k+1)}{(k+1)^2}$$

Thus, since $\frac{4(k+1)}{(k+2)}$ and $\frac{4^k}{k+1}$ are both positive, we have: $$\frac{4^{k+1}}{k+2}=\frac{4(k+1)}{(k+2)}\frac{4^k}{k+1} < \frac{4(k+1)}{(k+2)}\frac{(2k)!}{(k!)^2}<\frac{(2k+2)(2k+1)}{(k+1)^2}\frac{(2k)!}{(k!)^2} = \frac{(2k+2)!}{(((k+1)!)^2}$$

edited Sep 20 '17 at 16:05

answered Sep 20 '17 at 13:58

Jaroslaw Matlak

4,895

Why can you bring that other inequality to mine equality? – Anonymous196 Sep 20 '17 at 15:09
Why not? As you track the last chain of inequalities you'l see, that at each step I swap one factor with the bigger one (first $\frac{4^k}{k+1}$ with $\frac{(2k)!}{(k!)^2}$ - your inductive assumption, and then $\frac{4(k+1)}{k+2}$ with $\frac{(2k+2)(2k+1)}{(k+1)^2}$ - my inequality, that I've prooved for $k>1$). As a result I've obtained your inductive thesis. Are there some forbidden steps that I've missed? – Jaroslaw Matlak Sep 20 '17 at 15:48
Yes I've noticed that but how do you come up to prove that particular statement and use it in such a manner to prove my inequality? – Anonymous196 Sep 20 '17 at 15:51
I didn't even know you could prove this like that. I just thought of induction and nothing else. – Anonymous196 Sep 20 '17 at 15:52
Notice, that both sides of my inequality are factors, that are contained in your inductive thesis, but are missing in inductive assumption. I've checked, if they satisfy my inequality (track my proof from the end to the begining) and then prooved it. In my answer there is not a full inductive proof of your inequality - I've just prooved that the thesis is right if the assumption is true. – Jaroslaw Matlak Sep 20 '17 at 16:04

Prove that for $n \ge 2$ the follow inequality holds $\frac {4^n}{n+1} \lt \frac{(2n)!}{(n!)^2}$.

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