Please Be Zeta of Something

Calculus Level 3

n = 1 1 n 2 ( n + 1 ) ( n + 2 ) = π a b c d \sum_{n=1}^{\infty}\frac{1}{n^2(n+1)(n+2)} = \frac{\pi^a}{b} - \frac{c}{d}

The above sum is true for integers a , b , c , d a,b,c,d , where c , d c,d are coprime positive integers. Give your answer as a + b + c + d a+b+c+d .


The answer is 27.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

1 n 2 ( n + 1 ) ( n + 2 ) = 1 2 n 2 3 4 ( 1 n 1 n + 1 ) + 1 4 ( 1 n + 1 1 n + 2 ) \dfrac {1}{n^2(n+1)(n+2)}=\dfrac {1}{2n^2}-\dfrac {3}{4}(\frac{1}{n}-\frac{1}{n+1})+\dfrac {1}{4}(\frac{1}{n+1}-\frac{1}{n+2})

Therefore the sum is equal to 1 2 × π 2 6 3 4 × 1 + 1 4 × 1 2 = π 2 12 5 8 \dfrac {1}{2}\times \dfrac {π^2}{6}-\dfrac {3}{4}\times 1+\dfrac {1}{4}\times \dfrac {1}{2}=\dfrac {π^2}{12}-\dfrac {5}{8}

Hence a = 2 , b = 12 , c = 5 , d = 8 a=2,b=12,c=5,d=8 , and a + b + c + d = 27 a+b+c+d=\boxed {27} .

Brilliant ,just brilliant Mr.Alak .

Aruna Yumlembam - 12 months ago
Chew-Seong Cheong
Jun 17, 2020

S = n = 1 1 n 2 ( n + 1 ) ( n + 2 ) = n = 1 ( 1 n ( n + 1 ) ) ( 1 n ( n + 2 ) ) = n = 1 ( 1 n 1 n + 1 ) 1 2 ( 1 n 1 n + 2 ) = n = 1 1 2 ( 1 n 2 1 n ( n + 1 ) 1 n ( n + 2 ) + 1 ( n + 1 ) ( n + 2 ) ) = n = 1 1 2 ( 1 n 2 1 n + 1 n + 1 1 2 ( 1 n 1 n + 2 ) + 1 n + 1 1 n + 2 ) = n = 1 1 2 ( 1 n 2 3 2 n + 2 n + 1 1 2 ( n + 2 ) ) = 1 2 n = 1 1 n 2 3 4 n = 1 ( 1 n 1 n + 1 ) + 1 4 n = 1 ( 1 n + 1 1 n + 2 ) Riemann zeta function ζ ( s ) = k = 1 1 k s = 1 2 ζ ( 2 ) 3 4 1 + 1 4 1 2 and ζ ( 2 ) = π 2 6 = π 2 12 5 8 \begin{aligned} S & = \sum_{n=1}^\infty \frac 1{n^2(n+1)(n+2)} \\ & = \sum_{n=1}^\infty \left(\frac 1{n(n+1)}\right)\left(\frac 1{n(n+2)}\right) \\ & = \sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+1}\right)\cdot \frac 12 \left(\frac 1n - \frac 1{n+2}\right) \\ & = \sum_{n=1}^\infty \frac 12 \left(\frac 1{n^2} - \frac 1{n(n+1)} - \frac 1{n(n+2)} + \frac 1{(n+1)(n+2)} \right) \\ & = \sum_{n=1}^\infty \frac 12 \left(\frac 1{n^2} - \frac 1n + \frac 1{n+1} - \frac 12 \left(\frac 1n - \frac 1{n+2} \right) + \frac 1{n+1} - \frac 1{n+2} \right) \\ & = \sum_{n=1}^\infty \frac 12 \left(\frac 1{n^2} - \frac 3{2n} + \frac 2{n+1} - \frac 1{2(n+2)} \right) \\ & = \frac 12 \blue{\sum_{n=1}^\infty \frac 1{n^2}} - \frac 34 \sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+1}\right) + \frac 14 \sum_{n=1}^\infty \left(\frac 1{n+1} - \frac 1{n+2}\right) & \small \blue{\text{Riemann zeta function }\zeta (s) = \sum_{k=1}^\infty \frac 1{k^s}} \\ & = \frac 12 \blue{\zeta (2)} - \frac 34 \cdot 1 + \frac 14 \cdot \frac 12 & \small \blue{\text{and }\zeta (2) = \frac {\pi^2} 6} \\ & = \frac {\pi^2}{12} - \frac 58 \end{aligned}

Therefore a + b + c + d = 2 + 12 + 5 + 8 = 27 a+b+c+d = 2+12+5+8 = \boxed{27} .


Reference: Riemann zeta function

Another incredible solution,but there is a problem with zeta of 2 ,sir you are forgetting π 2 \pi^2

Aruna Yumlembam - 11 months, 4 weeks ago

Log in to reply

Thanks. How could I miss that.

Chew-Seong Cheong - 11 months, 4 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...