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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 (11/2)*(11/3)*...*(11/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)...)((n1)/n)=1/n. The limit is obviously 0.

October 3rd, 2018, 02:24 PM  #3  
Member Joined: Aug 2016 From: Romania Posts: 32 Thanks: 1  Quote:
(n1)*(n2)*(n3)!/n*(n1)*(n2)*(n3)!. 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  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552  Quote:
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 
Global Moderator Joined: May 2007 Posts: 6,835 Thanks: 733  You did not understand what I wrote. (n1)/n is the least term of the product. When it is concatenated, the product is (n1)!/n!.
Last edited by skipjack; October 4th, 2018 at 11:29 PM. 

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