4
$\begingroup$

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

$\endgroup$
2
  • 2
    $\begingroup$ You probably mean $\frac{2i-1}{2i}$? $\endgroup$ Nov 2, 2014 at 16:44
  • $\begingroup$ yes, thank you. $\endgroup$
    – Asinomás
    Nov 2, 2014 at 16:47

6 Answers 6

11
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ The $N$-th harmonic number: $$H_N=\sum_{k=1}^{N}\frac{1}{k}>\sum_{k=1}^{N}\log\frac{k+1}{k}=\log(N+1).$$ $\endgroup$ Nov 2, 2014 at 17:00
10
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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 $x_n=\frac{1}{2n}$.

$\endgroup$
1
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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 $A=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.