Simplify Multiplication of Fractions

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SUMMARY

The discussion focuses on simplifying the sum of a series of fractions involving factorials, specifically the expression $\dfrac{3}{3\times 4}+\dfrac{4}{3\times 4\times 5}+\cdots+\dfrac {99}{3\times 4\times 5\times \cdots\times 99\times 100}$. The series is transformed into a telescoping sum, leading to the conclusion that the total equals $2 \left(\dfrac{1}{3!}-\dfrac{1}{100!}\right)$. This simplification utilizes properties of factorials and series summation techniques.

PREREQUISITES
  • Understanding of factorial notation and operations
  • Familiarity with telescoping series
  • Basic knowledge of summation notation
  • Experience with algebraic manipulation of fractions
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  • Study the properties of telescoping series in detail
  • Explore advanced techniques in series summation
  • Learn about the applications of factorials in combinatorics
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Mathematicians, educators, students studying calculus or algebra, and anyone interested in advanced series manipulation techniques.

Albert1
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$\dfrac{3}{3\times 4}+\dfrac{4}{3\times 4\times 5}+\dfrac{5}{3\times 4\times 5\times 6}+\cdots+\dfrac {99}{3\times 4\times 5\times 6\times \cdots\times 99\times 100}$
 
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Re: count value

Albert said:
$\dfrac{3}{3\times 4}+\dfrac{4}{3\times 4\times 5}+\dfrac{5}{3\times 4\times 5\times 6}+ \cdots +\dfrac {99}{3\times 4\times 5\times 6\times \cdots \times 99\times 100}$

$=2 \left( \dfrac{3}{2 \times 3\times 4}+\dfrac{4}{2\times 3\times 4\times 5}+\dfrac{5}{2\times 3\times 4\times 5\times 6}+ \cdots +\dfrac {99}{2 \times 3\times 4\times 5\times 6\times \cdots \times 99\times 100} \right)$

$$=2 \sum^{100}_ {n=4} \frac{n-1}{n!}=2 \sum^{100}_ {n=4} \frac{1}{(n-1)!}-\frac{1}{n!} $$

The sum is telescoping

$$2\left( \dfrac{1}{3!} -\dfrac{1}{4!} + \dfrac{1}{4!} - \dfrac{1}{5!} + \frac{1}{5!}-\frac{1}{6!}+ \cdots +\dfrac {1}{99!} - \dfrac {1}{100!} \right)= 2 \left(\dfrac{1}{3!}-\dfrac{1}{100!} \right) $$
 

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