Big, ugly, but not so complicated

Algebra Level 4

n = 2 2017 1 n k = 2 n ( 1 + 1 k ) \large \displaystyle \sum_{n=2}^{2017} \dfrac {\frac{1}{n}}{\displaystyle \prod_{k=2}^{n} \left( 1+\frac{1}{k} \right)}

The above expression can be expressed in the form a b \dfrac {a}{b} , where a a and b b are coprime positive integers. Find a + b a+b .


The answer is 2017.

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3 solutions

Rishabh Jain
Feb 27, 2016

Let Q \mathfrak{Q} denote the given expression. k = 2 n ( 1 + 1 k ) = k = 2 n ( k + 1 k ) \Large \displaystyle \prod_{k=2}^{n} \left( 1+\frac{1}{k}\right)=\displaystyle\prod_{k=2}^{n}\left(\dfrac{k+1}{k}\right) = 2 × × × × n + 1 = n + 1 2 =\Large\dfrac{\color{#D61F06}{\not 3}}{2}\times \dfrac{\color{#0C6AC7}{\not 4}}{\color{#D61F06}{\not 3}}\times \dfrac{\color{#456461}{\not 5}}{\color{#0C6AC7}{\not 4}}\times\cdots \times \dfrac{n+1}{\color{#EC7300}{\not n}} =\dfrac{n+1}{2} Hence , Q = n = 2 2017 1 n k = 2 n ( 1 + 1 k ) \large \mathfrak{Q}=\displaystyle \sum_{n=2}^{2017} \dfrac {\frac{1}{n}}{\displaystyle \prod_{k=2}^{n}\left(1+\frac{1}{k}\right)} = n = 2 2017 2 n ( n + 1 ) \large =\displaystyle\sum_{n= 2}^{2017}\dfrac{2}{n(n+1)} = 2 n = 2 2017 ( 1 n 1 n + 1 ) \large ~=2\displaystyle\sum_{n=2}^{2017}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right) ( T e l e s c o p i c S e r i e s ) \large\color{#302B94}{\mathcal{(Telescopic ~Series)}} = 2 ( 1 2 1 2018 ) = 1008 1009 \Large =2\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)=\dfrac{1008}{1009} 1008 + 1009 = 2017 \Large \therefore~1008+1009=\huge\color{#456461}{\boxed{\color{#D61F06}{\boxed{\color{#007fff}{\textbf{2017}}}}}}

Every Problem I solve and when I come to the solutions I see only 2 words everytime on the top: Rishabh Cool.

Kushagra Sahni - 5 years, 3 months ago

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lol,very true @Rishabh Cool does write beautiful solutions,keep up the good work! :)

Hamza A - 5 years, 3 months ago

Exactly. The guys got something I have to say

Shreyash Rai - 5 years, 3 months ago

Do u use latex editor(which one?)?

Mehul Chaturvedi - 5 years, 3 months ago

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No... ............

Rishabh Jain - 5 years, 3 months ago
Chew-Seong Cheong
Feb 28, 2016

n = 2 2017 1 n k = 2 n ( 1 + 1 k ) = n = 2 2017 1 n k = 2 n ( k + 1 k ) = n = 2 2017 1 n ( n + 1 ) ! 2 n ! = n = 2 2017 2 n ! n ( n + 1 ) ! = 2 n = 2 2017 ( n 1 ) ! ( n + 1 ) ! = 2 n = 2 2017 ( n 1 ) ! ( n 1 ) ! ( 1 n ( n + 1 ) ) = 2 n = 2 2017 ( 1 n 1 n + 1 ) = 2 ( n = 2 2017 1 n n = 3 2018 1 n ) = 2 ( 1 2 1 2018 ) = 1 1 1009 = 1008 1009 \begin{aligned} \sum_{n=2}^{2017} \frac{\dfrac{1}{n}}{\displaystyle \prod_{k=2}^n\left(1+ \frac{1}{k} \right)} & = \sum_{n=2}^{2017} \frac{\dfrac{1}{n}}{\displaystyle \prod_{k=2}^n\left(\frac{k+1}{k} \right)} \\ & = \sum_{n=2}^{2017} \frac{\dfrac{1}{n}}{\dfrac{(n+1)!}{2n!}} \\ & = \sum_{n=2}^{2017} \frac{2n!}{n(n+1)!} \\ & = 2 \sum_{n=2}^{2017} \frac{(n-1)!}{(n+1)!} \\ & = 2 \sum_{n=2}^{2017} \frac{(n-1)!}{(n-1)!}\left(\frac{1}{n(n+1)}\right) \\ & = 2 \sum_{n=2}^{2017} \left(\frac{1}{n}-\frac{1}{n+1} \right) \\ & = 2 \left(\sum_{n=2}^{2017} \frac{1}{n}-\sum_{n=3}^{2018} \frac{1}{n} \right) \\ & = 2 \left(\frac{1}{2} - \frac{1}{2018} \right) \\ & = 1 - \frac{1}{1009} = \frac{1008}{1009} \end{aligned}

a + b = 1008 + 1009 = 2017 \Rightarrow a + b = 1008+1009 = \boxed{2017}

Nihar Mahajan
Feb 27, 2016

k = 2 n 1 + 1 k = k = 2 n k + 1 k = 3 2 × 4 3 × 5 4 × × n n 1 × n + 1 n = n + 1 2 \prod_{k=2}^{n} 1+\dfrac{1}{k}= \prod_{k=2}^{n}\dfrac{k+1}{k} = \dfrac{3}{2} \times \dfrac{4}{3} \times \dfrac{5}{4} \times \dots \times \dfrac{n}{n-1} \times \dfrac{n+1}{n} = \dfrac{n+1}{2}

Substituting this in given summation ,

n = 2 2017 1 n n + 1 2 = n = 2 2017 2 n ( n + 1 ) = 2 n = 2 2017 ( n + 1 ) n n ( n + 1 ) = 2 n = 2 2017 1 n 1 n + 1 = 2 [ 1 2 1 3 + 1 3 1 4 + 1 4 1 5 1 2017 + 1 2017 1 2018 ] = 2 [ 1 2 1 2018 ] = 2 × 1 2 [ 1 1 1009 ] = 1009 1 1009 = 1008 1009 \begin{aligned} \sum_{n=2}^{2017} \dfrac{\frac{1}{n}}{\frac{n+1}{2}} &= \sum_{n=2}^{2017} \dfrac{2}{n(n+1)} \\ &=2 \sum_{n=2}^{2017} \dfrac{(n+1) - n}{n(n+1)} \\ &= 2 \sum_{n=2}^{2017} \dfrac{1}{n} - \dfrac{1}{n+1} \\ &= 2 \left[\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4} - \dfrac{1}{5} \dots -\dfrac{1}{2017} + \dfrac{1}{2017}-\dfrac{1}{2018}\right] \\ &= 2 \left[\dfrac{1}{2}-\dfrac{1}{2018}\right] \\ &=2 \times \dfrac{1}{2} \left[1-\dfrac{1}{1009}\right] \\ &= \dfrac{1009-1}{1009} \\ &= \boxed{\dfrac{1008}{1009}} \end{aligned}

Hence a = 1008 , b = 1009 a + b = 1008 + 1009 = 2017 a=1008 \ , \ b=1009 \Rightarrow a+b=1008+1009=\boxed{2017} .

Moderator note:

Great explanation of the telescoping product and sum.

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