How to calculate complicated factorial

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In summary, the conversation revolves around evaluating a limit involving factorials. The initial attempt is to substitute infinity for n, but it is then explained that this is not a valid approach. The expert suggests multiplying the expression by $\displaystyle \begin{align*} \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \end{align*}$ and then expanding and simplifying to determine the limit as n approaches infinity. The final result is 1/4.
  • #1
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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  • #2
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?
 
  • #3
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.
 
  • #4
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 $\displaystyle \begin{align*} n \to \infty \end{align*}$.
 
  • #5
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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  • #6
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).
 
  • #7
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!
 

FAQ: How to calculate complicated factorial

How do I calculate a factorial?

A factorial is calculated by multiplying a number by all the numbers that come before it, starting from 1. For example, 5! (read as "5 factorial") is equal to 5 x 4 x 3 x 2 x 1 = 120.

What is the formula for calculating a factorial?

The formula for calculating a factorial is n! = n x (n-1) x (n-2) x ... x 2 x 1. In words, it means that the factorial of a number is equal to that number multiplied by all the numbers that come before it, until 1 is reached.

Can I calculate a factorial for non-integer numbers?

No, factorials are only defined for positive integers. This is because the factorial function involves multiplying numbers together, and it doesn't make sense to multiply a number by a non-whole number.

Is there a limit to the size of numbers that can be used in factorial calculations?

Technically, there is no limit to the size of numbers that can be used in factorial calculations. However, as the numbers get larger, they become more difficult to work with and may require special mathematical techniques or computer programs to calculate.

How do I calculate factorials for numbers with decimals or negative numbers?

Factorials are only defined for positive integers. To calculate factorials for numbers with decimals or negative numbers, you can use the Gamma function, which is a generalization of the factorial function. It is defined for all complex numbers except for negative integers.

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