Range of $N=\frac21\times\frac43\times\dots\times\frac{400}{399}$?

Question

Question:

If $N=\frac21\times\frac43\times\dots\times\frac{400}{399}$ then it lies between:
a) $50$ and $60$
b) $20$ and $30$
c) $30$ and $40$
d) $40$ and $50$

What I tried:

$M1$: Tried finding of someway to reduce the expression. Taking $2$ common from all the terms in numerator didn't help. Then writing terms as $(1+\frac1{2k-1})$ also didn't help.
$M2$: Multiplied first few terms to check if I get somewhere close to the answer as the latter terms are very small. But nope, first few terms didn't really give me a nice idea.

Please help.

Bonus it is involved in calculating $\int_0^{\pi}\cos(x)^{400}dx$ — zwim, Nov 19 '22 at 11:43
@RobertIsrael whoa! It's that easy to write! Thanks. Though not what I was looking for but helpful. — InanimateBeing, Nov 20 '22 at 15:25
You can also search for some similar solutions using approach0.xyz — Sil, Nov 20 '22 at 15:45

Sayan Dutta · Answer 1 · 2022-11-19T10:07:21.157

3

It is a well known fact that $$\frac 1{\sqrt{4n+2}}\le \frac{1.3.5\dots (2n-1)}{2.4.6\dots (2n)}\le \frac 1{\sqrt{3n+1}}$$ It can be proven by induction or by noticing the fact that $$\frac ab\le \frac{a+k}{b+k}$$ and then taking a clever product.

edited Nov 19 '22 at 10:07

answered Nov 19 '22 at 10:02

Sayan Dutta

8,831

Could you please tell the name of this well known fact? Or maybe provide some link to it so I can check it? – InanimateBeing Nov 19 '22 at 10:57
1

https://math.stackexchange.com/q/3004689/399263 – zwim Nov 19 '22 at 11:39
Thank you for your answer. +1. – InanimateBeing Nov 19 '22 at 12:38
@InanimateBeing you're welcome! You may also want to check out Gautschi's inequality – Sayan Dutta Nov 19 '22 at 13:46
1

The LHS can be improved to $1/(2\sqrt{n}),$ because $((2n-1)/(2n))^2 > (n - 1)/n.$ – Calum Gilhooley Nov 19 '22 at 15:12

NN2 · Answer 2 · 2022-11-25T03:00:20.487

2

$$N = \frac{\left(2\times4\times...\times400 \right)^2}{400!} =\frac{\left(2^{200}200! \right)^2}{400!}$$ Applying the Stirling's formula: $$n! \approx \sqrt{2\pi n}\left(\frac{n}{e} \right)^n$$ Then $$\color{red}{N} \approx\frac{\left(2^{200}\sqrt{2\pi \times200}\left(\frac{200}{e} \right)^{200} \right)^2}{\sqrt{2\pi \times400}\left(\frac{400}{e} \right)^{400}}=\sqrt{2\pi}\times 10 \color{red}{\approx25.07}$$

Then, we should choose $\color{red}{\mathbf{b}}$: $N$ lies between $20$ and $30$.

Q.E.D

edited Nov 25 '22 at 03:00

answered Nov 19 '22 at 09:32

NN2

15,892

1

Thank you for your answer. +1. Though not really what I was looking for but it's correct and every difficult (to calculate) term got cancelled. So... nice! :) – InanimateBeing Nov 19 '22 at 09:40
1

@CalumGilhooley I delete the complaint. Thank you for your understanding and your advice. :) – NN2 Nov 25 '22 at 03:03

score 1 · Answer 3 · answered Nov 20 '22 at 15:16

Here's what I got from a friend:

First we find the lower limit of $N$.
We know that $\frac21>\frac32,\frac43>\frac54,\dots\frac{398}{397}>\frac{399}{398} \ \text{and }\frac{400}{399}>1$
Let $X=\left(\frac32\right)\left(\frac54\right)\dots\left(\frac{399}{398}\right)(1)$
$NX=400$. Also, $N>X$
So, $\bf N>20$

Now we'll find the upper limit of $N$.
$N^2=\left(\frac21\right)^2\left(\frac43\right)^2\left(\frac65\right)^2\dots\left(\frac{400}{399}\right)^2$
Let $Y=\left(\frac21\right)^2\left(\frac32\times\frac43\right)\left(\frac54\times\frac65\right)\dots\left(\frac{399}{398}\times\frac{400}{399}\right)$
$N^2<Y$. Also, $Y=(2)(400)\Rightarrow N^2<(2)(400)$
So, $\bf N<20\sqrt2\sim28.2$

Range of $N=\frac21\times\frac43\times\dots\times\frac{400}{399}$?

3 Answers3

Thus, $\bf N$ lies between $\bf 20$ and $\bf 30$.