How to calculate complicated factorial

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SUMMARY

This discussion focuses on calculating complicated factorial limits, specifically evaluating the expression involving factorials as n approaches infinity. The key formula discussed is |(n + 1)^2 / ((2n + 2)(2n + 1))|, which simplifies the limit evaluation. Participants emphasize the importance of not substituting infinity directly and recommend expanding brackets and dividing by the highest power of n to identify negligible terms. The correct limit is confirmed to be 1/4.

PREREQUISITES
  • Understanding of factorial notation and properties
  • Familiarity with limits in calculus
  • Ability to manipulate algebraic expressions
  • Knowledge of indeterminate forms in calculus
NEXT STEPS
  • Learn about evaluating limits involving factorials
  • Study the concept of indeterminate forms and L'Hôpital's Rule
  • Explore advanced techniques for simplifying complex algebraic expressions
  • Investigate the behavior of sequences and series as they approach infinity
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus and advanced algebra, will benefit from this discussion on factorial limits and evaluation techniques.

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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