Proof $\frac{1}{2}\cdot\frac{3}{4}\cdot\frac {5}{6}\dots$ is zero

I would like a proof that $$\lim\limits_{ n\to \infty }\prod_{i=1}^n\frac{2i-1}{2i}= 0$$

It seems reasonable, although the terms approach $1$

Thank you in advance

• You probably mean $\frac{2i-1}{2i}$? – Thomas Andrews Nov 2 '14 at 16:44
• yes, thank you. – Jorge Fernández Hidalgo Nov 2 '14 at 16:47

I give an elementary proof by proving the following inequality. $$\prod_{i=1}^n\frac{2i-1}{2i}<\frac{1}{\sqrt{2n+1}}.$$ This can be proved by $$\frac{2i-1}{2i}<\frac{\sqrt{2i-1}}{\sqrt{2i+1}},$$ which is easy.

$$\prod_{n=1}^{N}\left(1-\frac{1}{2n}\right)=\exp\sum_{n=1}^{N}\log\left(1-\frac{1}{2n}\right)\leq\exp\left(-\frac{H_N}{2}\right)\leq\frac{1}{\sqrt{N+1}}.$$

• The $N$-th harmonic number: $$H_N=\sum_{k=1}^{N}\frac{1}{k}>\sum_{k=1}^{N}\log\frac{k+1}{k}=\log(N+1).$$ – Jack D'Aurizio Nov 2 '14 at 17:00

Hint Take a log and use the inequality $\ln(1+x)\le x$ so $$\ln \prod_{k=1}^n\frac{2k-1}{2k}=\sum_{k=1}^n\ln \frac{2k-1}{2k}\le -H_n.$$

In general, if $0<x_n<1$ and $x_n\to 0$ and $\sum x_n$ diverges, then $$\lim_{N\to\infty} \prod_{n=1}^N (1-x_n) = 0$$

In this case $x_n=\frac{1}{2n}$.

I wonder if this approach is correct:

Define $X_n = \prod_{i=1}^n \frac{2i-1}{2i}$. Then $log(X_n) = log(1) - log(2) + log(3) - log(4) + \ldots + log(n-1) - log(n)$. Per definition it will always end with the terms $log(n-1) - log(n)$, thus we write

\begin{eqnarray} log(X_N) &=& \left(\vphantom{x^2}log(1) - log(2)\right) + \left(\vphantom{x^2}log(3) - log(4)\right) + \ldots + \left(\vphantom{x^2}log(n-1) - log(n)\right) \end{eqnarray}

and since $$1 < \frac{log(n-3) - log(n-2)}{log(n-1) - log(n)} \leq 2.41 \; \; \;$$

the terms will add up to $-\infty$ if $n \rightarrow \infty$ and by definition of the $log$

\begin{eqnarray} log(X_n) &=& -\infty\\ X_n &=& 0. \end{eqnarray}

Let \begin{align}\frac12\frac34\frac56\cdots&=A\\ \frac23\frac45\frac67\cdots&=B\end{align} Then note that $B\geq A\geq0$ but $AB=0$, so $A=0$.