I'm trying to find the limit: $\displaystyle\lim_{n\to\infty}\frac {1\cdot2\cdot3\cdot...\cdot n}{(n+1)(n+2)...(2n)}$

I thought of taking a pretty obvious binding from above expression: $\frac {n^n} {(n+1)^n}$ which is $n$ times the largest numerator and $n$ times the smallest denominator, but this limit isn't zero and it looks like the given does go to zero.

So maybe 'multiplying' $\frac 1 {n+1}$ by $1$, $n$ times is a good bound but it isn't very obvious that: $\frac {1\cdot2\cdot3\cdot...\cdot n}{(n+1)(n+2)...(2n)}\le \frac 1 {n+1}$ so why is it true?


The numerator has a term 1, so upper bound $n^{n-1}$ is acceptable.

Since you found denominator is at least $(n+1)^n$, you get $$ \frac{1\cdots n}{(n+1)\cdots 2n} \leq \frac{n^{n-1}}{(n+1)^n}\leq \frac1{n+1}.$$

  • $\begingroup$ Oh now I got it. Thanks. $\endgroup$ – shinzou Dec 8 '14 at 20:29

Write your left-hand side as

$${1 \over (n+1)} \times {2 \cdot 3 \cdot \cdots \cdot n \over (n+2)(n+3) \cdot \cdots \cdot 2n}$$

Now the first factor is $1/(n+1)$, and the second factor is obviously less than 1 (since the numerator is smaller than the denominator). So the product is less than $1/(n+1)$.

This throws out a lot of information, but your bound is very weak - as has been observed, your function is approximately $\sqrt{\pi n}/4^n$, which is much smaller than $1/(n+1)$.

Edited to add: we can rewrite the product as $(n!)^2/(2n)!$. Using Stirling's approximation $n! \approx \sqrt{2\pi n} (n/e)^n$ you get

$$(n!)^2/(2n)! \approx { (2\pi n) (n/e)^{2n} \over \sqrt{4 \pi n} (2n/e)^{2n}}$$

and you can simplify that to get the approximation $\sqrt{\pi n}/4^n$.

  • $\begingroup$ How do you get to $\sqrt{\pi n}/4^n$? $\endgroup$ – shinzou Dec 8 '14 at 20:58
  • $\begingroup$ See my edits to answer that question. $\endgroup$ – Michael Lugo Dec 8 '14 at 21:08

Try to express your term with binomial coefficients!

  • $\begingroup$ A better approximation for your term is $\frac{\sqrt{\pi n}}{4^n}$ using Stirling's formula. See also here: en.wikipedia.org/wiki/Central_binomial_coefficient $\endgroup$ – Tintarn Dec 8 '14 at 20:26
  • $\begingroup$ Isn't it kinda complicated to get to that expression on your own? $\endgroup$ – shinzou Dec 8 '14 at 20:33
  • $\begingroup$ Of course it is. I just wanted to point out that there is actually a much stronger estimation for your term. $\endgroup$ – Tintarn Dec 9 '14 at 13:35

Hint: $\binom{2n}{k}$ is unimodal by moving k with maximum at $\binom{2n}{n}$. See also that $n+1\leq 2n=\binom{2n}{1}$.

  • $\begingroup$ How is unimodality relevant here? $\endgroup$ – shinzou Dec 8 '14 at 20:31
  • $\begingroup$ because $\binom{2n}{k}\leq \binom{2n}{n}$ for every possible $k$, in particular for $k=1$. $\endgroup$ – Phicar Dec 8 '14 at 20:32

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