Need help understanding this step in simplifying a limit equation

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SUMMARY

The discussion centers on the conversion of a limit equation as presented in a specific paper. The key transformation involves recognizing that the continued product from 1 to l+1 results in ((l+1)!)^2. Additionally, the terms in the numerator can be simplified by dividing by the denominator, which includes a sequence of odd integers squared. The manipulation of the denominator's terms, particularly the inclusion of (2j-1)/2 and (2j+1)/2, is crucial for achieving the desired formula (5). Furthermore, the term √π is squared and repositioned to the numerator, while the algebraic simplification of 1/(2l+3)/2 to 2/(2l+3) is also highlighted.

PREREQUISITES
  • Understanding of factorial notation and properties, specifically ((l+1)!)^2
  • Familiarity with continued products and their applications in limit equations
  • Knowledge of algebraic manipulation involving square roots and fractions
  • Basic comprehension of sequences, particularly odd integer sequences
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  • Study the properties of factorials and their role in limit equations
  • Research continued products and their significance in mathematical analysis
  • Explore algebraic techniques for simplifying complex fractions and square roots
  • Investigate sequences of odd integers and their applications in mathematical proofs
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Mathematicians, students studying advanced calculus, and anyone involved in mathematical research or analysis of limit equations.

mody mody
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I read a paper and i can't understand how can we make the conversion (that attached in photo)
i mean how i get formula (5) from previous one
 

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I think I figured this one out. The continued product of each ## j ## from ## 1 ## to ## l+1 ## is the ## ((l+1)!)^2 ##. The 2's in the numerator actually can be divided in the denominator. The sequence of 1*3*5*7...(squared) in the denominator takes on two forms depending whether the last term is ## (2l+1)/2 ## or includes the ## (l+\frac{3}{2}) ## term, thereby the continued product of ## (2j-1)/2 ## and ## (2j+1)/2 ## in the denominator. The ## \sqrt{\pi} ## in the denominator gets squared and moved to the numerator of the other side, and the ## \frac{1}{2} ## is a simple algebraic term because ## \frac{1}{(2l+3)/2}=\frac{2}{2l+3} ##.
 

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