300 Followers Question

Just for a change, take log on both the sides and make pairs(let the value of expression be S)

$\displaystyle \ln(S) = \sum_{n = 1}^{\infty } \left( \ln(n+1) - \ln(n) \right) + \sum_{n=1}^{\infty } \left( \ln(n+2) - \ln(n+3) \right)$

Now they both are telescoping series, the first one will converge to 0 and the second one will converge to $\ln(3)$

$\therefore \ln S = ln3 \implies \boxed{\boxed{S = 3}}$

Lol I solved this orally

We know that

$\displaystyle \prod_{n=1}^{\infty} \dfrac{(n+1)(n+2)}{n(n+3)} = \prod_{n=1}^{\infty} (n+1) \times \prod_{n=1}^{\infty} (n+2) \times \prod_{n=1}^{\infty} \dfrac{1}{n} \times \prod_{n=1}^{\infty} \dfrac{1}{(n+3)}$

Let's deal with the first product .

$\displaystyle \prod_{n=1}^{\infty} (n+1) = \lim_{n \rightarrow \infty} (1+1)(1+2)(1+3)\dots(1+n)$

Let's use the Gauss form of Gamma Function , which is

$\displaystyle \Gamma(x) = \lim_{n \rightarrow \infty} \dfrac{n^{x} n!}{x(x+1)(x+2) \dots (x+n)}$

$\therefore$ We have ,

$\displaystyle x \Gamma(x) = \lim_{n \rightarrow \infty} \dfrac{n^{x} n!}{(x+1)(x+2) \dots (x+n)}$

Which gives us ,

$\displaystyle \prod_{n=1}^{\infty} (x+n) = \dfrac{n^{x}n!}{x\Gamma(x)}$

$\therefore$ We get the following ,

$\displaystyle \prod_{n=1}^{\infty} (1+n) = \dfrac{n^{1}n!}{\Gamma(1)} \\ \displaystyle \prod_{n=1}^{\infty} (2+n) = \dfrac{n^{2}n!}{2\cdot\Gamma(2)} \\ \displaystyle \prod_{n=1}^{\infty} (3+n) = \dfrac{n^{3}n!}{3\cdot\Gamma(3)}$

And we also know that $\prod_{r=1}^{n} \frac{1}{r} = \frac{1}{n!}$

Now all that's left is some multiplication and division ,

$\displaystyle \prod_{n=1}^{\infty} \dfrac{(n+1)(n+2)}{n(n+3)}$

$= \dfrac{n^{1}n!}{\Gamma(1)} \times \dfrac{n^{2}n!}{2\cdot \Gamma(2)} \times \dfrac{1}{n!} \times \dfrac{3\cdot\Gamma(3)}{n^{3}n!}$

$= \dfrac{3\cdot \Gamma(3)}{2\cdot \Gamma(2)}$

$= 3$

$\prod _{ n=1 }^{ k }{ \frac { (n+1)(n+2) }{ { n(n+3) } } } =\frac { k!\times \frac { k! }{ 2! } }{ k!\times \frac { k! }{ 3! } } =\frac { \frac { 1 }{ 2! } }{ \frac { 1 }{ 3! } } =3$

Solved this in my head too... Nice problem!

Basically, $n=1$ , and the only thing that increases in each new fraction is what's being summed to $n$ in each of the factors of its numerator and denominator. But I will still use $n$ , instead of $1$ , until the very end of the solution, because I think it's easier to visualize it this way:

Simply predicting how many times each number will appear in the numerator and in the denominator:

$\displaystyle\prod_{n=1}^\infty \dfrac{(n+1)(n+2)}{n(n+3)} = \dfrac{1}{n} \cdot \dfrac{n+1}{n+1} \cdot \dfrac{(n+2)^2}{n+2} \cdot \left(\dfrac{n+3}{n+3}\right)^2 \cdot \left(\dfrac{n+4}{n+4}\right)^2 \cdot \left(\dfrac{n+5}{n+5}\right)^2 \cdots$

$= \dfrac{1}{n} \cdot 1 \cdot \dfrac{n+2}{1} \cdot 1^2 \cdot 1^2 \cdot 1^2 \cdots = \dfrac{n+2}{n} = \dfrac{1+2}{1} = \boxed{3}$

As we can see denominator is larger than numerator the term converges

So after some point we would be very near to answer as the term will converge to 1

Here is my C code

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23

#include<stdio.h>

int main()

{

    int i;

    double ans=1;   

    for(i=1;i<10000;i++)

    {

        ans = ans * (i+1) * (i+2);

        ans = (ans / (i))/(i+3);

    }

    printf("%d\n",(int)(ans+0.5));

}

The line (int)(ans+0.5) is simply to round - off to upper bound

Firstly, write out the series and simplify as much as possible: $\frac{(n+1)(n+2)}{n(n+3)} \times\frac{(n+2)(n+3)}{(n+1)(n+4)} \times\frac{(n+3)(n+4)}{(n+2)(n+5)} \times\ ...$ Now we start with n=1 so we are left with: $3 \times\lim_{n \rightarrow\infty } \frac{n-2}{n} = 3 \times\lim_{n \rightarrow\infty } (1 - \frac{2}{n} ) = \boxed{3}$