How can I find the limit of the following sequence $\frac{1}{2} . \frac {3}{4} ....... \frac {2n - 1}{2n} $?

Question

How can I find the limit of the following sequence:

$$\frac{1}{2} . \frac {3}{4} ....... \frac {2n - 1}{2n} $$

The problem for me is that they are multiplied and not added, also I can see that the numerator is the set of odd numbers and the denominator is the set of even numbers but then what?

Answer 1

Write $a_n$ for $\frac 1 2 \cdot \frac 3 4 \cdots \frac{2n-1}{2n}$ and $b_n$ for $\frac 2 3 \cdot \frac 4 5 \cdots \frac{2n}{2n+1}$. Then clearly $a_n < b_n$; but $a_n \cdot b_n = \frac 1 {2n+1}$ so $a_n \cdot b_n \to 0$ as $n \to \infty$. And $a_n^2 < a_n \cdot b_n$, so it must also be that $a_n \to 0$. That's your limit.

Answer 2

Apply the log we get \begin{equation} \log \prod_{n=1}^N \frac{2n-1}{2n}=\sum_{n=1}^N \log \left(1-\frac{1}{2n}\right)\le -\frac12 \sum_{n=1}^N\frac 1n, \end{equation} because $\log (1 + x) \leq x $ for all $x > -1$. So \begin{equation} \prod_{n=1}^N \frac{2n-1}{2n} \leq e^{-\frac12 \sum_{n=1}^N\frac 1n} \end{equation} But $\sum_{n=1}^N\frac 1n \rightarrow \infty$ by the $p-$test. So, \begin{equation} \lim_{N \rightarrow \infty} \prod_{n=1}^N \frac{2n-1}{2n} \leq \lim_{N \rightarrow \infty} e^{-\frac12 \sum_{n=1}^N\frac 1n} = 0 \end{equation} By the sandwich theorem \begin{equation} \lim_{N \rightarrow \infty} \prod_{n=1}^N \frac{2n-1}{2n} = 0 \end{equation}

Answer 3

Let $S_n=\prod_{j=1}^n\frac {2j-1}{2j}.$ We have $$\frac {1}{S_n}=\prod_{j=1}^n(1+\frac {1}{2j-1})\geq \sum_{j=1}^n\frac {1}{2j-1}.$$ E.g. if $a,b,c$ are non-negative then $(1+a)(1+b)=1+a+b+ab\geq 1+a+b,$ and $(1+a)(1+b)(1+c)\geq$ $ (1+a+b)(1+c)=1+a+b+c+(a+b)c\geq 1+a+b+c.$

So $\frac {1}{S_n}\to \infty.$ So $S_n\to 0.$

Remarks: Two useful elementary tools:

(1). If each $a_n\geq 0$ then $\sum_{n\in \Bbb N}a_n$ converges iff $\prod_{n\in \Bbb N}(1+a_n)$ converges.

(2). If $0\leq a_n<1$ for each $n$ then $\sum_{n\in \Bbb N}a_n$ converges iff $\prod_{n\in \Bbb N}(1-a_n)\ne 0.$

In this Q, we may apply (2) with $a_n=1/2n$ to get $S_n\to 0$ directly, or, as I did above, apply (1) to the lower bound $\sum_{j=1}^n(1+1/(2n-1))$ for $1/S_n$ to get $1/S_n\to \infty.$