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Prove that $$ \lim_{n\to\infty}{\left(\frac12\cdot\frac34\cdot\frac56\cdot\ldots\cdot\frac{2n-1}{2n}\right)}=0. $$

Transforming it to factorial obviously doesn't help at all, so I've noted $A_n$ as above product and noticed $1/2<2/3$, $3/4<4/5$, ..., $(2n-1)/(2n)<(2n)/(2n+1)$, so $A_n<1/\sqrt{2n+1}$.

Now it's kind of obvious that $A_n$ approached $0$ as $n$ approaches infinity, but I'm not sure about formality of this proof. Is it safe to conclude that $A_n\to0^+ \text{when } n\to\infty$ from $A_n>0 \land A_n<1/\sqrt{2n+1}=0^+ \text{when } n\to\infty$?

Later
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Lazar Ljubenović
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    The way you've written it now, it actually tends to 5/16. – Joe Z. Mar 12 '13 at 14:37
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    Yeah I forgot $\cdots$, gonna fix it now. – Lazar Ljubenović Mar 12 '13 at 14:37
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    Since nobody mentioned the squeeze theorem, I will do it. – J.H. Mar 12 '13 at 15:14
  • I wouldnt like to open another topic, so I will ask it here: Can someone explain why $A_n < \frac{1}{\sqrt{2n+1}}$? – Giiovanna Oct 13 '14 at 17:24
  • Related: https://math.stackexchange.com/questions/1899857/mathematical-problem-induction-frac12-cdot-frac34-cdots-frac2n-12n-frac/, https://math.stackexchange.com/questions/431234/induction-and-convergence-of-an-inequality-frac1-cdot3-cdot5-cdots2n-12?rq=1, https://math.stackexchange.com/questions/1627933/prove-frac1-cdot-3-cdots2n-12-cdot-4-cdots2n-frac1-sqrt2n1?rq=1 – StubbornAtom Jan 24 '18 at 19:03
  • You could $a_n=\frac{1}{2}\times \frac{2}{3}\times \dots \times \frac{2n-1}{2n}$ and then write $-\ln(a_n)=\sum_{j=1}^n\ln\big(1+\frac{1}{2j-1}\big)$. In light of this, we see $a_n\longrightarrow 0$ if and only if $\sum_{n=1}^\infty\ln\big(1+\frac{1}{2n-1}\big)=+\infty$.

    You can use limit comparison test to show $\sum_{n=1}^\infty\ln\big(1+\frac{1}{2n-1}\big)=+\infty$ by comparing this infinite series to $\sum_{n=1}^{\infty}\frac{1}{2n-1}$.

     – Matthew H. Aug 25 '23 at 13:00

2 Answers2

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Yes, since $\displaystyle\frac1{\sqrt{2n+1}}\,\to 0$, so we have $$0<A_n<\frac1{\sqrt{2n+1}}\to 0\,.$$

Berci
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If you put some effort, you can have the following identity

$$ \prod_{m=1}^{n}\frac{2m-1}{2m}= {\frac { \left( \frac{1}{2} \right)^{n+1}{2}^{n+1}\left( n-\frac{1}{2} \right) !}{n! \,\sqrt {\pi }}}.$$

Using Stirling approximation for $ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n $ and taking limit as $n$ goes to infinity, the desired result follows.

Mhenni Benghorbal
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