1. ## Limit of Sequence

$\displaystyle \lim_{n \to \infty } \frac{(n!)^2}{(2n)!}$

2. ## Re: Limit of Sequence

Originally Posted by roshanhero
$\displaystyle \lim_{n \to \infty } \frac{(n!)^2}{(2n)!}$
Here is a hint: $\displaystyle (2n)!=2^n(n!)[(2n-1)(2n-3)\cdots3\cdot 1].$

3. ## Re: Limit of Sequence

Originally Posted by Plato
Here is a hint: $\displaystyle (2n)!=2^n(n!)[(2n-1)(2n-3)\cdots3\cdot 1].$
$\displaystyle \frac{n!}{2^{n}[(2n-1)(2n-3)....3.1]}$
I have reached upto here Now, What should I do?

4. ## Re: Limit of Sequence

Originally Posted by roshanhero
$\displaystyle \lim_{n \to \infty } \frac{(n!)^2}{(2n)!}$
I will denote $\displaystyle x=\frac{(n!)^2}{(2n)!}$, and take natural logarithm from both sides.

See what happens...

5. ## Re: Limit of Sequence

Let $\displaystyle u_n :=\frac{(n!)^2}{(2n)!}\neq 0$. Since $\displaystyle \lim_{n\to\infty}\frac{u_{n+1}}{u_n}=\frac 14$, the limit of the sequence $\displaystyle \{u_n\}$ is $\displaystyle 0$. Maybe it will be more interesting to look at $\displaystyle \lim_{n\to +\infty}\frac{4^n(n!)^2}{(2n)!}$.

6. ## Re: Limit of Sequence

Another tricky way is to test the convergence of the corresponding series
The series $\displaystyle \sum \dfrac{(n!)^2}{(2n)!}$ converges by the Ratio Test.
Thus, $\displaystyle \lim_{n\to\infty} \dfrac{(n!)^2}{(2n)!} = 0$ (Theorem).

7. ## Re: Limit of Sequence

I simply have to find the limit as n tends to infinity, I don't have to use any methods of tests of convergence here

8. ## Re: Limit of Sequence

According to my way...

$\displaystyle x=\frac{(n!)^2}{(2n)!}$

$\displaystyle \ln{x}=\ln{\frac{(n!)^2}{(2n)!}}$

$\displaystyle \ln{\frac{(n!)^2}{(2n)!}} =$

$\displaystyle \ln{(n!)^2}-\ln{(2n)!}=$

$\displaystyle 2\ln{(n!)}-\ln{(2n)!}=$

$\displaystyle 2[\ln{1}+\ln{2}+\ln{3}+...+\ln{n}]-[\ln{1}+\ln{2}+\ln{3}+...+\ln{n}+\ln{(n+1)...\ln{2n }}]=$

$\displaystyle \ln{1}+\ln{2}+\ln{3}+...+\ln{n}-[\ln{(n+1)+...+\ln{2n}}]=$

$\displaystyle \ln{\frac{1}{n+1}}+\ln{\frac{2}{n+2}}+\ln{\frac{3} {n+3}}+...+\ln{\frac{n}{2n}}=$

$\displaystyle \ln{\frac{n!}{(n+1)(n+2)(n+3)...(2n)}}$

$\displaystyle \ln{x}=\ln{\frac{n!}{(n+1)(n+2)(n+3)...(2n)}}$

$\displaystyle x=\frac{n!}{(n+1)(n+2)(n+3)...(2n)}$

$\displaystyle \frac{2^n}{(2n)^n}<\frac{n!}{(n+1)(n+2)(n+3)...(2n )}\leq \frac{1}{2^n}$

Now just apply the sandwich rule.

9. ## Re: Limit of Sequence

Originally Posted by roshanhero
$\displaystyle \lim_{n \to \infty } \frac{(n!)^2}{(2n)!}$
Note that $\displaystyle \frac{(2n)!}{(n!)^2}$ is the Binomial Coefficient $\displaystyle \binom{2n}{n}$, which is a positive integer.

We have the equality: $\displaystyle \binom{2n}{n}\cdot\frac{2n+1}{n+1}\cdot\frac{2n+2} {n+1}=\binom{2(n+1)}{n+1}$.

But $\displaystyle \frac{2n+1}{n+1}\cdot\frac{2n+2}{n+1}>1$ and therefore $\displaystyle \binom{2n}{n}<\binom{2(n+1)}{n+1}$. The strict inequaliy shows that the integers $\displaystyle \binom{2n}{n}$ tend to infinity as $\displaystyle n$ gets larger.

Then $\displaystyle \lim_{n \to \infty}\frac{(n!)^2}{(2n)!}=\lim_{n \to \infty}\frac{1}{\binom{2n}{n}}=0$.

To AlsoSprachZarathusra, there's no need to take logarithms to get the last two lines!

10. ## Re: Limit of Sequence

for large n, $\displaystyle \frac{(n!)^{2}}{(2n)!} \approx \frac{2 \pi n (\frac{n}{e})^{2n}}{2 \sqrt{\pi n} (\frac{2n}{e})^{2n}}$

$\displaystyle = \frac{\sqrt{\pi n}}{2^{2n}}$

11. ## Re: Limit of Sequence

Originally Posted by roshanhero
$\displaystyle \lim_{n \to \infty } \frac{(n!)^2}{(2n)!}$

melese reminded me...

$\displaystyle 4^n=(1+1)^{2n}=\sum^{2n}_{k=0}\binom{2n}{k}<(2n+1) \binom{2n}{n}$

Therefor:

$\displaystyle \binom{2n}{n}>\frac{4^n}{2n+1}$

Or:

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

12. ## Re: Limit of Sequence

Originally Posted by Also sprach Zarathustra
According to my way...

$\displaystyle x=\frac{(n!)^2}{(2n)!}$

$\displaystyle \ln{x}=\ln{\frac{(n!)^2}{(2n)!}}$

$\displaystyle \ln{\frac{(n!)^2}{(2n)!}} =$

$\displaystyle \ln{(n!)^2}-\ln{(2n)!}=$

$\displaystyle 2\ln{(n!)}-\ln{(2n)!}=$

$\displaystyle 2[\ln{1}+\ln{2}+\ln{3}+...+\ln{n}]-[\ln{1}+\ln{2}+\ln{3}+...+\ln{n}+\ln{(n+1)...\ln{2n }}]=$

$\displaystyle \ln{1}+\ln{2}+\ln{3}+...+\ln{n}-[\ln{(n+1)+...+\ln{2n}}]=$

$\displaystyle \ln{\frac{1}{n+1}}+\ln{\frac{2}{n+2}}+\ln{\frac{3} {n+3}}+...+\ln{\frac{n}{2n}}=$

$\displaystyle \ln{\frac{n!}{(n+1)(n+2)(n+3)...(2n)}}$

$\displaystyle \ln{x}=\ln{\frac{n!}{(n+1)(n+2)(n+3)...(2n)}}$

$\displaystyle x=\frac{n!}{(n+1)(n+2)(n+3)...(2n)}$

$\displaystyle \frac{2^n}{(2n)^n}<\frac{n!}{(n+1)(n+2)(n+3)...(2n )}\leq \frac{1}{2^n}$

Now just apply the sandwich rule.
How did u get $\displaystyle \frac{2^n}{(2n)^n}<\frac{n!}{(n+1)(n+2)(n+3)...(2n )}\leq \frac{1}{2^n}$ as lower and upper bounds?

13. ## Re: Limit of Sequence

Originally Posted by roshanhero
How did u get $\displaystyle \frac{2^n}{(2n)^n}<\frac{n!}{(n+1)(n+2)(n+3)...(2n )}\leq \frac{1}{2^n}$ as lower and upper bounds?
$\displaystyle \frac{n!}{(n+1)\cdot(n+2)\cdot(n+3)\cdot...\cdot(2 n)}=\frac{1\cdot 2\cdot 3 \cdot ...\cdot n}{(n+1)\cdot(n+2)\cdot(n+3)\cdot...\cdot(2n)}=$

$\displaystyle \frac{1}{n+1} \cdot \frac{2}{n+2} \cdot \frac{3}{n+3} \cdot... \cdot\frac{n}{2n}\leq \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot... \cdot\frac{1}{2}=\frac{1}{2^n}$

$\displaystyle \frac{ n! }{ (n+1) \cdot (n+2) \cdot (n+3) \cdot ...\cdot (2n) } = \frac{ 1 \cdot 2 \cdot 3 \cdot ... \cdot n }{ (n+1) \cdot (n+2) \cdot (n+3) \cdot ...\cdot (2n) } > \frac{ 2 \cdot 2 \cdot 2 \cdot ... \cdot 2 }{ (2n) \cdot (2n) \cdot (2n) \cdot ... \cdot (2n) } = \frac{ 2^n }{ (2n)^n }$

14. ## Re: Limit of Sequence

Hello, roshanhero!

$\displaystyle \lim_{n \to \infty } \frac{(n!)^2}{(2n)!}$
$\displaystyle \text{Let: }\:R \;=\;\frac{a_{n+1}}{a_n}$

$\displaystyle R \;=\;\frac{[(n+1)!]^2}{[2(n+1)]!}\cdot\frac{(2n)!}{(n!)^2} \;=\;\frac{(n+1)!(n+1)!}{(n!)(n!)}\cdot\frac{(2n)! }{(2n+2)(2n+1)(2n)!}$

. . $\displaystyle =\;\frac{(n+1)(n+1)}{(2n+2)(2n+1)}$

Divide top and bottom by $\displaystyle n^2: }\:\frac{\left(1 + \frac{1}{n}\right)\left(1 + \frac{1}{n}\right)}{\left(2 + \frac{2}{n}\right)\left(2 + \frac{1}{n}\right)}$

. . $\displaystyle \lim_{n\to\infty}R \;=\;\lim_{n\to\infty} \frac{\left(1 + \frac{1}{n}\right)\left(1 + \frac{1}{n}\right)}{\left(2 + \frac{2}{n}\right)\left(2 + \frac{1}{n}\right)} \;=\;\frac{(1+0)(1+0)}{(2+0)(2+0)} \;=\;\frac{1}{4}$

As $\displaystyle n$ gets larger, the sequence approximates
. . a geometric sequence with common ratio $\displaystyle \tfrac{1}{4}$

Therefore, the limit of the sequence is zero.