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October 3rd, 2018, 01:59 PM   #1
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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.
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October 3rd, 2018, 03:00 PM   #2
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The expression with n terms is (1/2)(2/3)(3/4)...)((n-1)/n)=1/n. The limit is obviously 0.
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October 3rd, 2018, 03:24 PM   #3
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Quote:
Originally Posted by mathman View Post
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 5th, 2018 at 12:28 AM.
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October 3rd, 2018, 07:52 PM   #4
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Is it okay how I solve it? I wanted to know where the result 1/n appeared from.

Last edited by skipjack; October 5th, 2018 at 12:28 AM.
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October 3rd, 2018, 08:21 PM   #5
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Quote:
Originally Posted by alex77 View Post
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 5th, 2018 at 12:29 AM.
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October 3rd, 2018, 09:13 PM   #6
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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 $n + 1$.)
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October 4th, 2018, 08:26 PM   #7
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Quote:
Originally Posted by alex77 View Post
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 5th, 2018 at 12:29 AM.
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