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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?

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    Suppose it has a limit, call it $L$. Now, consider $\ln L$. Now it is a summation and you can use your usual techniques. – JMoravitz Sep 06 '18 at 01:42
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    @Batominovski - perhaps, but the linked threads (the answers) use more work to do more than simply answer the question about the limit. I didn't see the very elementary answer to the present question in those threads. –  Sep 06 '18 at 02:02
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    @mathguy The second answer by Ahmad Bazzi does not use much knowledge. Plus, there are two more links that are linked to that thread. About that you "didn't see the very elementary answer," why should you expect "very elementary answer"? Just because a question looks simple does not mean it will have a "very elementary" solution. In fact, this question happens many times, but I am too lazy to find all old links. – Batominovski Sep 06 '18 at 02:07
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    @Batominovski - I didn't assume something, I asserted it. The problem does have a very elementary solution, I just posted one. –  Sep 06 '18 at 02:09
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    @Batominovski thanks for the diss ;) – Ahmad Bazzi Sep 06 '18 at 02:10
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    @mathguy As I said, this question popped up many times. This is another thread: https://math.stackexchange.com/questions/2402508/convergence-point-of-frac-prod-k-1n-2k-1-prod-k-1n-2k/2402623#2402623. It has an answer with exactly the same solution that you gave. (I just realized that I also gave exactly the same solution in two threads.) – Batominovski Sep 06 '18 at 02:11
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    lol - OK, I see it. Then, if you want to suggest a question is a duplicate of an older one, you should perhaps spend just a little longer to find the right "original". –  Sep 06 '18 at 02:13
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    @mathguy What? That is an absurd statement. Any link with an answer to this question is a "right original." The method used in an answer is irrelevant. And I didn't need to give the oldest link to prove that this question is a duplicate. This question is a duplicate no matter which old link I happen to use to justify my claim. – Batominovski Sep 06 '18 at 02:14
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    @JavaMan - how can a question asking to compute a limit be a duplicate of a question asking to compute a closed form of the general term? As I showed, finding the limit is obviously a different (much easier) question, since it has a solution that doesn't help in any way with finding a closed form. "This question can be answered as a trivial corollary of another question" doesn't mean "this" question is a duplicate of the other one. –  Sep 06 '18 at 02:51
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    Duplicates include: https://math.stackexchange.com/questions/2886753/limit-of-the-product-prod-k-1n-left-frac2k-12k-right, https://math.stackexchange.com/questions/1586773/evaluate-prod-frac-2k-1-2k, https://math.stackexchange.com/questions/93001/closed-form-expression-for-the-product-prod-limits-k-1n-left1-frac1, https://math.stackexchange.com/questions/2402508/convergence-point-of-frac-prod-k-1n-2k-1-prod-k-1n-2k?noredirect=1&lq=1 OP was too lazy to look for duplicates, so question should be closed. Your answer is nice, but answer a different version of the question if you want the points. – JavaMan Sep 06 '18 at 02:56
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    @mathguy Moreover, the question I link when I voted to close actually evaluates the sum, so yes it is a duplicate whichever way you want to slice it. If there three newer questions each asking how to compute $\int_1^2 x^2dx$, and yet there was an older post showing how to evaluate $\int_1^t x^2dx$ for all $t$, wouldn't you agree that the older answer is sufficient? – JavaMan Sep 06 '18 at 03:13
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    @JavaMan - In your example, I wouldn't agree if the definite integral had a much simpler way to be computed. Change your example to $\int_{-1}^{1} x^3 dx$ vs $\int_{-1}^{t}\dots$. The former question has a trivial answer which has nothing to do with the more general question. Neither is a duplicate of the other. –  Sep 06 '18 at 03:17

3 Answers3

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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.

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}

JavaMan
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Ahmad Bazzi
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1

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 $1/S_n\to \infty.$

DanielWainfleet
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