$\prod_{n=1}^{\infty} \dfrac{(n+1)(n+2)}{n(n+3)} = \, ?$
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You don't even need to use logarithm. You can use the technique of telescoping product. Note that the product can be rewritten as,
$\displaystyle\prod_{n=1}^\infty \frac{(n+1)(n+2)}{n(n+3)}=\frac{\left(\displaystyle\prod_{n=2}^\infty n\right)\cdot\left(\displaystyle\prod_{n=3}^\infty n\right)}{\left(\displaystyle\prod_{n=1}^\infty n\right)\cdot\left(\displaystyle\prod_{n=4}^\infty n\right)}=\frac{2\times 3^2}{2\times 3}\cdot \left(\displaystyle\prod_{n=4}^\infty \frac{n^2}{n^2}\right)=\boxed{3}$
The product converges because the denominator grows faster than the numerator.
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Great solution yar...
Its just that I am more comfortable with log :)
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Well yes, everybody has their own preferences. :)
Greatly solved. Nice
you are BRILLIANT bro......
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I'm nowhere near brilliant, dude! I'm just an ordinary guy with somewhat of an average knowledge in math.
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@Prasun Biswas – i am glad to follow a guy like you............
Nice solution :) This one makes mine look so dumb .
+1 !
This was indeed an oral problem.
Nice solution.... Did in a similar way..
Figure these two out separately: (n+1)/n and (n+2)/(n+3). You have R+1 and 3/R+3= 3 × R+1/R+3 Since R is "infinite" R+1/R+3 = 1 So answer is 3
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 ,
$\displaystyle \prod_{n=1}^{\infty} \dfrac{(n+1)(n+2)}{n(n+3)}$
$= \dfrac{n^{1}n!}{\Gamma(1)} \times \dfrac{n^{2}n!}{2\cdot \Gamma(2)} \times \dfrac{1}{n!} \times \dfrac{3\cdot\Gamma(3)}{n^{3}n!}$
$= \dfrac{3\cdot \Gamma(3)}{2\cdot \Gamma(2)}$
$= 3$
Awesome technique ! thanks for sharing it @Azhaghu Roopesh M
Woah! You really damn overthought it!
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Nope!! It's just that I remember these identities so I solved it in 4 steps(but 4 is still longer) , I just wanted to make it clear to the others , but I guess I failed !
Also I created this question using the Gauss form , I never realised that you could bash this question so easily . But BEWARE , I will release a tougher version of this question :P
$\prod _{ n=1 }^{ k }{ \frac { (n+1)(n+2) }{ { n(n+3) } } } =\frac { k!\times \frac { k! }{ 2! } }{ k!\times \frac { k! }{ 3! } } =\frac { \frac { 1 }{ 2! } }{ \frac { 1 }{ 3! } } =3$
Solved this in my head too... Nice problem!
I can see a big error in your solution. You're considering a finite upper limit $k$ for the product. In that case, your value should be dependent on $k$ .
The main error you made in your solution was not noticing the fact that,
$\prod_{n=1}^k (n+m)=\frac{(k+m)!}{m!}\neq \frac{k!}{m!}~\textrm{for a finite }k\textrm{ and some constant }m$
Key point: When you shift the index of a sum or product where the limits are finite, the shifting is the same for both lower and upper limits because the number of terms of the sum/product is always the same.
Thanks $\ddot\smile$
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
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The line (int)(ans+0.5) is simply to round  off to upper bound
Hi Deep !
Type out your code using the following format (without the "") and write your code between C and the "```" marks .
"```C
```"
To get the output as :
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Thnxx Azhaghu Roopesh M
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: $3 \times\lim_{n \rightarrow\infty } \frac{n2}{n} = 3 \times\lim_{n \rightarrow\infty } (1  \frac{2}{n} ) = \boxed{3}$
Simplicity is beauty :)
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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