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 October 3rd, 2018, 12:59 PM #1 Member   Joined: Aug 2016 From: Romania Posts: 32 Thanks: 1 Problem with a limit which I do not understand How do I solve this? lim (1-1/2)*(1-1/3)*...*(1-1/x)=0 Also in my textbook I got the answer x->+inf 1/x. October 3rd, 2018, 02:00 PM #2 Global Moderator   Joined: May 2007 Posts: 6,835 Thanks: 733 The expression with n terms is (1/2)(2/3)(3/4)...)((n-1)/n)=1/n. The limit is obviously 0. October 3rd, 2018, 02:24 PM   #3
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Quote:
 Originally Posted by mathman The expression with n terms is (1/2)(2/3)(3/4)...)((n-1)/n)=1/n. The limit is obviously 0.
With a bit of correction, n-1 is in fact (n-1)! and n is n!. (n-1)!/n! is equal to
(n-1)*(n-2)*(n-3)!/n*(n-1)*(n-2)*(n-3)!. Now I understand the result.

Last edited by skipjack; October 4th, 2018 at 11:28 PM. October 3rd, 2018, 06:52 PM #4 Member   Joined: Aug 2016 From: Romania Posts: 32 Thanks: 1 Is it okay how I solve it? I wanted to know where the result 1/n appeared from. Last edited by skipjack; October 4th, 2018 at 11:28 PM. October 3rd, 2018, 07:21 PM   #5
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Quote:
 Originally Posted by alex77 Is it okay how I solve it? I wanted to know where the result 1/n appeared from.
The result comes intuitively comes from pattern recognition and formally from mathematical induction.

Intuitive

$n= 2 \implies \displaystyle \left ( \prod_{j=2}^2 1 - \dfrac{1}{j} \right ) = 1 - \dfrac{1}{2} = \dfrac{1}{2} = \dfrac{1}{n}.$

$n= 3 \implies \displaystyle \left ( \prod_{j=2}^3 1 - \dfrac{1}{j} \right ) = \left ( \prod_{j=2}^2 1 - \dfrac{1}{j} \right ) * \left ( 1 - \dfrac{1}{3} \right ) = \dfrac{1}{2} * \dfrac{2}{3} = \dfrac{1}{3} = \dfrac{1}{n}.$

Now, if you need to be formal, it should be obvious how to prove by weak mathematical induction that

$n \in \mathbb Z \text { and } n \ge 2 \implies \displaystyle \left ( \prod_{j=2}^n 1 - \dfrac{1}{j} \right ) = \dfrac{1}{n}.$

Last edited by skipjack; October 4th, 2018 at 11:29 PM. October 3rd, 2018, 08:13 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,030 Thanks: 2260 For $n$ terms (where $n \small\geqslant$ 1), $\displaystyle \prod_{k=1}^n\left(1 - \frac{1}{k + 1}\right) = \prod_{k=1}^n\frac{k}{k + 1} = \frac{n!}{(n + 1)!} = \frac{n!}{n!(n + 1)} = \frac{1}{n + 1}$, which tends to 0 as $n \to \infty$. (The original problem used $x$ instead of $n + 1$.) October 4th, 2018, 07:26 PM   #7
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Quote:
 Originally Posted by alex77 With a bit of correction, n-1 is in fact (n-1)! and n is n!. (n-1)!/n! is equal to (n-1)*(n-2)*(n-3)!/n*(n-1)*(n-2)*(n-3)!. Now I understand the result.
You did not understand what I wrote. (n-1)/n is the least term of the product. When it is concatenated, the product is (n-1)!/n!.

Last edited by skipjack; October 4th, 2018 at 11:29 PM. Tags limit, problem, understand Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post SMARTYPANTS Algebra 12 July 30th, 2018 05:48 PM Kallaste Probability and Statistics 4 April 19th, 2017 01:34 AM Angelwngs26 Elementary Math 4 April 10th, 2016 08:38 PM blackhype Calculus 3 March 24th, 2015 04:35 PM drunkelf22 Elementary Math 10 February 1st, 2015 09:22 AM

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