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It is already solved here at Math.stackexchange, but we haven't learned Stirling's approximation (at our school), so can it be solved using only squeeze theorem? $$\lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$$

My attempt,

Let $y_n = \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}$, we see that $ \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6\cdot ...\cdot 2n} > \frac{1 \cdot 2 \cdot 2 \cdot ... \cdot 2}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n} = \frac{1}{1 \cdot 2 \cdots .. \cdot (n-1)\cdot2n} = x_n$

$$\lim_{n \to \infty}{x_n} = 0 $$ Now I just need to find a sequnce $z_n>y_n$, so, $\lim_{n \to \infty} z_n = 0$.

Gjekaks
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2 Answers2

4

Hint multiply both numerator,denominator by $2.4.6....2n$ so you get $$\frac{(2n)!}{4^n(n!)^2}$$ where i get $4^n$ by taking $2$ common so we can write it as $$\frac{{2n\choose n}}{4^n}$$ thus we know $4^n\geq {2n\choose n}$. So limit is $0$.

Archis Welankar
  • 15,910
3

Let $$ b_n=\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots2n}. $$ We shall show inductively that $$ \frac{2}{\sqrt{2n+1}}>b_n. $$ For $n=1$ it clearly holds. Assume that $$ \frac{2}{\sqrt{2k+1}}>\frac{1\cdot 3\cdots (2k-1)}{2\cdot 4\cdots2k}. $$ Then $$ \frac{2}{\sqrt{2k+1}}\cdot\frac{2k+1}{2k+2}>\frac{1\cdot 3\cdots (2k-1)\cdot(2k+1)}{2\cdot 4\cdots(2k)\cdot(2k+2)}. $$ But $$ \sqrt{\frac{2k+3}{2k+1}}=\sqrt{1+\frac{2}{2k+1}}<1+\frac{1}{2k+1}=\frac{2k+2}{2k+1} $$

and hence $$ \frac{2}{\sqrt{2k+3}}>\frac{2}{\sqrt{2k+1}}\cdot\frac{2k+1}{2k+2}>b_{k+1}. $$

Another way to do it is by showing that $$ a_n=\frac{2\cdots 4\cdots 2n}{1\cdots 3\cdots (2n-1)}\to\infty. $$ Now, using the fact that, $$ x\in[0,1]\Longrightarrow \log(1+x)=\int_0^x\frac{dt}{1+t}\ge \frac{1}{2}\int_0^x\frac{dt}{1+t}=\frac{x}{2}, $$ we obtain $$ \log a_n=\sum_{k=1}^n\log \left(\frac{2k}{2k-1}\right)=\sum_{k=1}^n\log \left(1+\frac{1}{2k-1}\right)\ge \frac{1}{2}\sum_{k=1}^n \frac{1}{2k-1}. $$ It only remains to show that $$ 1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\to\infty. $$

Yiorgos S. Smyrlis
  • 83,933