How to calculate complicated factorial

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Discussion Overview

The discussion revolves around the calculation of a complicated factorial expression and the evaluation of its limit as \( n \) approaches infinity. Participants explore the mathematical steps involved in simplifying the expression and addressing the concept of substituting infinity in limit evaluations.

Discussion Character

  • Mathematical reasoning, Technical explanation, Debate/contested

Main Points Raised

  • Some participants present a factorial expression involving \( (n + 1)! \) and \( (2n + 2)! \) and seek guidance on how to calculate it.
  • There is a discussion about the proper method for evaluating limits, with some participants emphasizing that substituting infinity directly is not valid.
  • One participant suggests expanding the expression and multiplying by \( \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \) to clarify the behavior of terms as \( n \) approaches infinity.
  • Another participant claims to have found a limit value of \( \frac{1}{4} \) and questions the necessity of multiplying by \( \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \).
  • Responses indicate that dividing by the highest power of \( n \) helps identify negligible terms in the limit evaluation.

Areas of Agreement / Disagreement

Participants express differing views on the validity of substituting infinity in limit calculations, with some asserting it is incorrect. There is no consensus on the best approach to evaluate the limit, as participants propose various methods and reasoning.

Contextual Notes

Limitations include potential misunderstandings regarding the treatment of infinity in mathematical expressions and the need for clarity in the steps taken to evaluate limits involving factorials.

aruwin
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Hello.
I know what factorial means but how do I calculate this? Could someone explain to me on how to do it?
 

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aruwin said:
Hello.
I know what factorial means but how do I calculate this? Could someone explain to me on how to do it?

$\displaystyle \begin{align*} \left| \frac{\frac{ \left[ \left( n + 1 \right) ! \right] ^2 }{ \left( 2n + 2 \right) !} }{ \frac{ \left( n ! \right) ^2 }{ \left( 2n \right) ! } } \right| &= \left| \frac{\left[ \left( n + 1 \right) ! \right] ^2 }{ \left( 2n + 2 \right) ! } \cdot \frac{ \left( 2n \right) !}{\left( n! \right) ^2} \right| \\ &= \left| \frac{ \left[ \left( n + 1 \right) n! \right] ^2 \left( 2n \right) ! }{ \left( 2n + 2 \right) \left( 2n + 1 \right) \left( 2n \right) ! \left( n! \right) ^2 } \right| \\ &= \left| \frac{ \left( n + 1 \right) ^2 \left( n! \right) ^2 }{ \left( 2n + 2 \right) \left( 2n + 1 \right) \left( n ! \right) ^2 } \right| \\ &= \left| \frac{ \left( n + 1 \right) ^2}{\left( 2n + 2 \right) \left( 2n + 1 \right) } \right| \end{align*}$

Can you evaluate the limit now?
 
Prove It said:
$\displaystyle \begin{align*} \left| \frac{\frac{ \left[ \left( n + 1 \right) ! \right] ^2 }{ \left( 2n + 2 \right) !} }{ \frac{ \left( n ! \right) ^2 }{ \left( 2n \right) ! } } \right| &= \left| \frac{\left[ \left( n + 1 \right) ! \right] ^2 }{ \left( 2n + 2 \right) ! } \cdot \frac{ \left( 2n \right) !}{\left( n! \right) ^2} \right| \\ &= \left| \frac{ \left[ \left( n + 1 \right) n! \right] ^2 \left( 2n \right) ! }{ \left( 2n + 2 \right) \left( 2n + 1 \right) \left( 2n \right) ! \left( n! \right) ^2 } \right| \\ &= \left| \frac{ \left( n + 1 \right) ^2 \left( n! \right) ^2 }{ \left( 2n + 2 \right) \left( 2n + 1 \right) \left( n ! \right) ^2 } \right| \\ &= \left| \frac{ \left( n + 1 \right) ^2}{\left( 2n + 2 \right) \left( 2n + 1 \right) } \right| \end{align*}$

Can you evaluate the limit now?

Sorry, I am not sure how to evaluate the limit. I just know that n should be substituted with infinity.
 
aruwin said:
Sorry, I am not sure how to evaluate the limit. I just know that n should be substituted with infinity.

NO! You NEVER "substitute infinity", as infinity is NOT a number. Besides, $\displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$ is a meaningless indeterminate expression.

My advice would now be to expand out all the brackets, and then multiply by $\displaystyle \begin{align*} \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \end{align*}$. Once you have done this, you should be able to see what happens to each term as $\displaystyle \begin{align*} n \to \infty \end{align*}$.
 
Prove It said:
NO! You NEVER "substitute infinity", as infinity is NOT a number. Besides, $\displaystyle \begin{align*} \frac{\infty}{\infty} \end{align*}$ is a meaningless indeterminate expression.

My advice would now be to expand out all the brackets, and then multiply by $\displaystyle \begin{align*} \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \end{align*}$. Once you have done this, you should be able to see what happens to each term as $\displaystyle \begin{align*} n \to \infty \end{align*}$.

I got 1/4. One more question, why do we have to multiply by $\displaystyle \begin{align*} \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \end{align*}$ ?
 

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$$$$
aruwin said:
I got 1/4. One more question, why do we have to multiply by $\displaystyle \begin{align*} \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \end{align*}$ ?

1/4 is correct. I think you've answered your own question - when you divide by the highest power of n, you can see what the "negligible" terms are (i.e. the ones that go to 0).
 
Prove It said:
1/4 is correct. I think you've answered your own question - when you divide by the highest power of n, you can see what the "negligible" terms are (i.e. the ones that go to 0).

Thank you for your explanation!
 

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