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Algebra Level 3

1 2 ! + 2 3 ! + 3 4 ! + ⋯ + n − 1 n ! = ? \large \dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+\cdots +\dfrac{n-1}{n!} = \, ?

Notation :
! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 8! = 1\times2\times3\times\cdots\times8 .

( n + 1 ) ! n ! \frac{(n+1)!}{n!} 1 − 1 n ! 1-\frac{1}{n!} 1 ( n − 1 ) ! n ! \frac{(n-1)!}{n!}

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Margaret Zheng by Margaret Zheng , 20, USA

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2 solutions

Rishabh Jain
Mar 13, 2016

M = ∑ r = 2 n ( r − 1 ) r ! \Large \mathfrak{M}=\displaystyle\sum_{r=2}^n\dfrac{(r-1)}{r!} = ∑ r = 2 n ( 1 ( r − 1 ) ! − 1 r ! ) \Large =\displaystyle\sum_{r=2}^n\left(\dfrac{1}{(r-1)!}-\dfrac{1}{r!}\right) ( A T e l e s c o p i c S e r i e s ) \large\mathbf{\color{#0C6AC7}{(A~Telescopic~Series)}} ( 1 − 1 2 ! ) + ( 1 2 ! − 1 3 ! ) + ( 1 3 ! − 1 4 ) + ⋯ + ( 1 ( n − 1 ) ! − 1 n ! ) \large \left(1-\cancel{\color{#D61F06}{\dfrac{1}{2!}}}\right)+\left(\cancel{\color{#D61F06}{\dfrac{1}{2!}}}-\cancel{\color{#456461}{\dfrac{1}{3!}}}\right)+\left(\cancel{\color{#456461}{\dfrac{1}{3!}}}-\cancel{\color{#EC7300}{\dfrac{1}{4}}}\right)+ \cdots+\left(\cancel{\color{skyblue}{\dfrac{1}{(n-1)!}}}-\dfrac{1}{n!}\right) = 1 − 1 n ! \huge =1-\dfrac{1}{n!}

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Aditya Sky
Mar 24, 2016

Say given sum is S S . Clearly, S = ∑ m = 1 n − 1 m ( m + 1 ) ! S = \large\sum_{m=1}^{n-1}\frac{m}{(m+1)!} = ∑ m = 1 n − 1 m + 1 − 1 ( m + 1 ) ! = ∑ m = 1 n − 1 1 ( m ) ! − 1 ( m + 1 ) ! = \large\sum_{m=1}^{n-1}\frac{m +1 -1 }{(m+1)!} = \large\sum_{m=1}^{n-1}\frac{1}{(m)!} - \frac{1}{(m+1)!} = ( 1 1 ! − 1 2 ! ) + ( 1 2 ! − 1 3 ! ) + ( 1 3 ! − 1 4 ! ) + ⋅ ⋅ ⋅ + ( 1 ( n − 2 ) ! − 1 ( n − 1 ) ! ) + ( 1 ( n − 1 ) ! − 1 ( n ) ! ) = \left(\frac{1}{1!}\,-\, \frac{1}{2!} \right) \,+\, \left(\frac{1}{2!}\,-\, \frac{1}{3!} \right) \,+\, \left(\frac{1}{3!}\,-\, \frac{1}{4!} \right) \,+\, \cdot \cdot \cdot \,+\,\left(\frac{1}{(n-2)!}\,-\, \frac{1}{(n-1)!} \right) \,+\, \left(\frac{1}{(n-1)!}\,-\, \frac{1}{(n)!} \right) Hence, S = 1 − 1 n ! S\,=\,1\,-\ \frac{1}{n!}

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