Question

Another limit of sequence-

Answer 1

\[\lim_{n \rightarrow \infty} (n!)^{2}/(2n)!\]

Answer 2

well you can simplify that fraction a bit:
\[\frac{n!*n!}{n!*(n+1)(n+2)\ldots(2n)} = \frac{n!}{(n+1)(n+2)\ldots(2n)}\]

Answer 3

There are n items on the top and bottom, the product on the bottom is going to be bigger for sure. So the limit is 0? (i dont know, just asking)

Answer 4

the limit is zero...but you need more justification

Answer 5

Even I am stuck here......

Answer 6

pretty easy really ;)

Answer 7

Intuition says 0, because each of the factors in the denominator will be larger than any of the factors in the numerator for any value of n.

Answer 8

is this the same as:
\[\frac{1}{2nCn}\]

2nCn = \[\left(\begin{matrix}2n \\ n\end{matrix}\right)\]

Answer 9

factor out a \[2^n\] from the bottom

Answer 10

im not seeing that at all >.< from the initial problem or my reduced fraction?

Answer 11

Please write it serially guys, I am confused here

Answer 12

i mean, i see that half the terms are going to be even. is that where the 2s are coming from?

Answer 13

Different interpretation to the solution really!

Answer 14

oh oh ohoh ohoh oh oh i see it lol

Answer 15

\[\frac{n!}{(n+1)(n+2)\ldots(2n-1)(2n)}\]

\[=\frac{1\cdot 2\cdots (n-1)n}{{(\frac{n+1}{2})(\frac{n+2}{2})\ldots(\frac{2n-1}{2})(n)}}\frac{1}{2^n}\]

Answer 16

\[\frac{1\cdot 2\cdots (n-1)n}{{(\frac{n+1}{2})(\frac{n+2}{2})\ldots(\frac{2n-1}{2})(n)}}<1\]

Answer 17

I feeling messed up here.....LOL

Answer 18

match up the terms...clearly less than one

Answer 19

thus..

\[\frac{n!}{(n+1)(n+2)\ldots(2n-1)(2n)}<\frac{1}{2^n}\]

Answer 20

So, what's the limit here?

Answer 21

\[\frac{1}{2^n}\to 0\]

as \[n\to\infty\]

Answer 22

Oh Thanks ..............Hate these sequence

Answer 23

by the squeeze theorem :)

Answer 24

un fact is goes to 0 so damned fast that
\[\sum_0^{\infty} \frac{n!}{(n+1)(n+2)\ldots(2n)}<2\]

Answer 25

that is really what i just proved

Answer 26

Now satellite made me confused again......

Answer 27

ignore me

Answer 28

\[\sum_{n=0}^{\infty}\frac{1}{2^n}=2\]

Answer 29

right. just thought i would mention that not only is the limit 0, but it goes to zero lickedy split

Answer 30

Thanks guys...........Much appreciated. Now, I will try to understand it again.....

Answer 31

Here's a simpler way to see it, I think.

Like you guys said, it is simplified to
n!
__________________
(n+1)(n+2)...(n+n)

Rewrite this as
1 2 3 n
____ *____*____*...*____
n+1 n+2 n+3 n+n

Each denominator is just the numerator plus n, so each denominator is greater than the numerator which makes each term less than 1. This lets us consider the first term as an upper bound, since the whole sequence is just that term multiplied by some number of fractions less than 1.

1/(n+1) is an upper bound for the sequence, and 1/(n+1) approaches 0 as n approaches infinity.

The sequence is always positive, so it's always greater than 0. Therefore 0 is a lower bound.

By squeeze theorem, the limit is 0.