Sum up all of them! (1)

Calculus Level 3

A = n = 0 ( n 0 ) n ! + n = 1 ( n 1 ) n ! + n = 2 ( n 2 ) n ! + \large A = \sum_{n=0}^\infty \dfrac{ \binom n0}{n!} + \sum_{n=1}^\infty \dfrac{ \binom n1}{n!} + \sum_{n=2}^\infty \dfrac{ \binom n2}{n!} + \cdots

Given that A A has a closed form. Find 1000 A \lfloor 1000A \rfloor .

Notation : ( M N ) \dbinom MN denotes the binomial coefficient , ( M N ) = M ! N ! ( M N ) ! \dbinom MN = \dfrac{M!}{N!(M-N)!} .


This is part of set Sum up all of them!


The answer is 7389.

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3 solutions

Ishan Singh
May 27, 2016

Note that ( n r ) = 0 r > n \dbinom{n}{r} = 0 \ \forall \ r > n

n = 0 r = 0 ( n r ) 1 n ! = n = 0 r = 0 n ( n r ) 1 n ! \displaystyle \implies \sum_{n=0}^{\infty} \sum_{r=0}^{\infty} \dbinom{n}{r} \dfrac{1}{n!} = \sum_{n=0}^{\infty} \sum_{r=0}^{n} \dbinom{n}{r} \dfrac{1}{n!}

= n = 0 2 n n ! \displaystyle = \sum_{n=0}^{\infty} \dfrac{2^n}{n!}

= e 2 = \boxed{e^2}

Yes also correct! +1!

Joel Yip - 5 years ago

It's quite funny how we posted the three solutions at around the same time!

Joel Yip - 5 years ago
Chew-Seong Cheong
May 27, 2016

A = n = 0 ( n 0 ) n ! + n = 0 ( n 1 ) n ! + n = 0 ( n 2 ) n ! + n = 0 ( n 3 ) n ! + = n = 0 1 n ! + n = 0 n n ! + n = 0 n ( n 1 ) 2 ! n ! + n = 0 n ( n 1 ) ( n 2 ) 3 ! n ! + = n = 0 1 0 ! n ! + n = 1 1 1 ! ( n 1 ) ! + n = 2 1 2 ! ( n 2 ) ! + n = 3 1 3 ! ( n 3 ) ! + = 1 1 ! n = 0 1 n ! + 1 2 ! n = 0 1 n ! + 1 0 ! n = 0 1 n ! + 1 3 ! n = 0 1 n ! + = n = 0 1 n ! n = 0 1 n ! = ( n = 0 1 n ! ) 2 = e 2 \begin{aligned} A & = \sum_{n=0}^\infty \frac{\begin{pmatrix} n \\ 0 \end{pmatrix}}{n!} + \sum_{n=0}^\infty \frac{\begin{pmatrix} n \\ 1 \end{pmatrix}}{n!} + \sum_{n=0}^\infty \frac{\begin{pmatrix} n \\ 2 \end{pmatrix}}{n!} + \sum_{n=0}^\infty \frac{\begin{pmatrix} n \\ 3 \end{pmatrix}}{n!} + \cdots \\ & = \sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{n}{n!} + \sum_{n=0}^\infty \frac{n(n-1)}{2!n!} + \sum_{n=0}^\infty \frac{n(n-1)(n-2)}{3!n!} + \cdots \\ & = \sum_{n=0}^\infty \frac{1}{0!n!} + \sum_{n=1}^\infty \frac{1}{1!(n-1)!} + \sum_{n=2}^\infty \frac{1}{2!(n-2)!} + \sum_{n=3}^\infty \frac{1}{3!(n-3)!} + \cdots \\ & = \frac{1}{1!} \sum_{n=0}^\infty \frac{1}{n!} + \frac{1}{2!} \sum_{n=0}^\infty \frac{1}{n!} + \frac{1}{0!} \sum_{n=0}^\infty \frac{1}{n!} + \frac{1}{3!} \sum_{n=0}^\infty \frac{1}{n!} + \cdots \\ & = \sum_{n=0}^\infty \frac{1}{n!} \sum_{n=0}^\infty \frac{1}{n!} \\ & = \left( \sum_{n=0}^\infty \frac{1}{n!}\right)^2 \\ & = e^2 \end{aligned}

1000 A = 1000 e 2 = 7389 \implies \lfloor 1000A \rfloor = \lfloor 1000e^2 \rfloor = \boxed{7389}

Yes correct! +1!

Joel Yip - 5 years ago
Joel Yip
May 27, 2016

Firstly, I would like to mention that if n < m \displaystyle n<m , ( n m ) = 0 \displaystyle \left( \begin{matrix} n \\ m \end{matrix} \right) =0

Also, 1 ( 1 ) ! , 1 ( 2 ) ! , 1 ( 3 ) ! , = 0 \displaystyle \frac { 1 }{ \left( -1 \right) ! } ,\frac { 1 }{ \left( -2 \right) ! } ,\frac { 1 }{ \left( -3 \right) ! } ,\cdots =0

n = 0 ( n 0 ) n ! + n = 0 ( n 1 ) n ! + n = 0 ( n 2 ) n ! + = n = 0 ( n 0 ) + ( n 1 ) + ( n 2 ) + n ! = ( 0 0 ) + ( 0 1 ) + ( 0 2 ) + 0 ! + ( 1 0 ) + ( 1 1 ) + ( 1 2 ) + 1 ! + = ( 0 0 ) 0 ! + ( 1 0 ) + ( 1 1 ) 1 ! + ( 2 0 ) + ( 2 1 ) + ( 2 2 ) 2 ! + , a s i f n < m , ( n m ) = 0 = n = 0 k = 0 n ( n k ) n ! \displaystyle \sum _{ n=0 }^{ \infty }{ \frac { \left( \begin{matrix} n \\ 0 \end{matrix} \right) }{ n! } } +\sum _{ n=0 }^{ \infty }{ \frac { \left( \begin{matrix} n \\ 1 \end{matrix} \right) }{ n! } } +\sum _{ n=0 }^{ \infty }{ \frac { \left( \begin{matrix} n \\ 2 \end{matrix} \right) }{ n! } } +\cdots =\sum _{ n=0 }^{ \infty }{ \frac { \left( \begin{matrix} n \\ 0 \end{matrix} \right) +\left( \begin{matrix} n \\ 1 \end{matrix} \right) +\left( \begin{matrix} n \\ 2 \end{matrix} \right) +\cdots }{ n! } } \\ =\frac { \left( \begin{matrix} 0 \\ 0 \end{matrix} \right) +\left( \begin{matrix} 0 \\ 1 \end{matrix} \right) +\left( \begin{matrix} 0 \\ 2 \end{matrix} \right) +\cdots }{ 0! } +\frac { \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) +\left( \begin{matrix} 1 \\ 1 \end{matrix} \right) +\left( \begin{matrix} 1 \\ 2 \end{matrix} \right) +\cdots }{ 1! } +\cdots \\ =\frac { \left( \begin{matrix} 0 \\ 0 \end{matrix} \right) }{ 0! } +\frac { \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) +\left( \begin{matrix} 1 \\ 1 \end{matrix} \right) }{ 1! } +\frac { \left( \begin{matrix} 2 \\ 0 \end{matrix} \right) +\left( \begin{matrix} 2 \\ 1 \end{matrix} \right) +\left( \begin{matrix} 2 \\ 2 \end{matrix} \right) }{ 2! } +\cdots \quad \quad ,\quad as\quad if\quad n<m,\quad \left( \begin{matrix} n \\ m \end{matrix} \right) =0\\ =\sum _{ n=0 }^{ \infty }{ \frac { \sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) } }{ n! } }

In here , we see n = 0 k = 0 n ( n k ) n ! = n = 0 2 n n ! = e 2 \displaystyle \sum _{ n=0 }^{ \infty }{ \frac { \sum _{ k=0 }^{ n }{ \left( \begin{matrix} n \\ k \end{matrix} \right) } }{ n! } } =\sum _{ n=0 }^{ \infty }{ \frac { { 2 }^{ n } }{ n! } } \\ ={ e }^{ 2 }\\

OR

n = 0 ( n 0 ) n ! + n = 0 ( n 1 ) n ! + n = 0 ( n 2 ) n ! + = n = 0 n ! 0 ! × n ! n ! + n = 0 n ! 1 ! × ( n 1 ) ! n ! + n = 0 n ! 2 ! × ( n 2 ) ! n ! + = n = 0 1 0 ! × n ! + n = 0 1 1 ! × ( n 1 ) ! + n = 0 1 2 ! × ( n 2 ) ! + = n = 0 k = 1 1 k ! n ! , a s 1 ( 1 ) ! , 1 ( 2 ) ! , 1 ( 3 ) ! , = 0 \displaystyle \sum _{ n=0 }^{ \infty }{ \frac { \left( \begin{matrix} n \\ 0 \end{matrix} \right) }{ n! } } +\sum _{ n=0 }^{ \infty }{ \frac { \left( \begin{matrix} n \\ 1 \end{matrix} \right) }{ n! } } +\sum _{ n=0 }^{ \infty }{ \frac { \left( \begin{matrix} n \\ 2 \end{matrix} \right) }{ n! } } +\cdots =\sum _{ n=0 }^{ \infty }{ \frac { \frac { n! }{ 0!\times n! } }{ n! } } +\sum _{ n=0 }^{ \infty }{ \frac { \frac { n! }{ 1!\times \left( n-1 \right) ! } }{ n! } } +\sum _{ n=0 }^{ \infty }{ \frac { \frac { n! }{ 2!\times \left( n-2 \right) ! } }{ n! } } +\cdots \\ =\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ 0!\times n! } + } \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ 1!\times \left( n-1 \right) ! } } +\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ 2!\times \left( n-2 \right) ! } } +\cdots \\ =\sum _{ n=0 }^{ \infty }{ \frac { \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ k! } } }{ n! } } ,\quad as\quad \frac { 1 }{ \left( -1 \right) ! } ,\frac { 1 }{ \left( -2 \right) ! } ,\frac { 1 }{ \left( -3 \right) ! } ,\cdots =0

n = 0 e n ! = e 2 \sum _{ n=0 }^{ \infty }{ \frac { e }{ n! } } ={ e }^{ 2 }

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