I try to evaluate this limit:

$$\lim_{n\to+\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$$

I considered this inequality

$$\frac{1}{4n}\le\left [ \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)} \right]^2\le \frac{1}{2n+1}$$

and so

$$\lim_{n\to+\infty} \frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}=0$$

my questions are:

1)- how do I prove the inequality with the principle of induction?

2)- there is another way to solve this limit?

  • 4
    $\begingroup$ Note that you really only need the upper inequality since the limit is clearly nonegative. $\endgroup$ – JavaMan Dec 23 '11 at 20:42
  • 6
    $\begingroup$ "Solve" is the wrong word. You're trying to evaluate a limit. $\endgroup$ – Michael Hardy Dec 23 '11 at 20:43

I will only show the upper inequality as @JavaMan's comment shows. For $n=1$, it's only $\frac 14\leq\frac 13$, and if it's true for $n$ then \begin{align*}\left(\prod_{j=1}^{n+1}\frac{2j-1}{2j}\right)^2&=\left(\frac{2n+1}{2(n+1)}\right)^2\left(\prod_{j=1}^n\frac{2j-1}{2j}\right)^2\\ &\leq \left(\frac{2n+1}{2(n+1)}\right)^2\frac 1{2n+1}\\ &=\frac{2n+1}{4(n+1)^2}, \end{align*} which is $\leq \frac 1{2n+3}$ since $\frac{(2n+1)(2n+3)}{4(n+1)^2}=\frac{4n^2+5n+3}{4n^2+8n+4}\leq 1$.

An other way: put $a_n:=\prod_{j=1}^n\frac{2j-1}{2j}$. Then by $\ln(1+x)\leq x$ for $x\geq -1$ we have $$\ln a_n =\sum_{j=1}^n\ln\left(1-\frac 1{2j}\right)\leq -\frac 12\sum_{j=1}^n\frac 1j,$$ so $a_n=\exp\left(-\frac 12\sum_{j=1}^n\frac 1j\right)$ and we can conclude since the harmonic series diverges.


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