A relevant summation

Calculus Level 5

1 ( 1 ) ( 2 ) ( 3 ) + 2 ( 4 ) ( 5 ) ( 6 ) + 3 ( 7 ) ( 8 ) ( 9 ) + 4 ( 10 ) ( 11 ) ( 12 ) + … \large\displaystyle\dfrac{1}{(1)(2)(3)}+\dfrac{2}{(4)(5)(6)}+\dfrac{3}{(7)(8)(9)}+\dfrac{4}{(10)(11)(12)}+\ldots

Let S \large S denote the value of series above. Find the value of ⌊ 10000 S ⌋ \large\lfloor10000S\rfloor .


The answer is 2015.

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Ikkyu San by Ikkyu San , 23, Thailand

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3 solutions

Akiva Weinberger
May 17, 2015

Note that ∫ 0 1 x n − 1 d ⁡ ⁣ x = 1 n \displaystyle\int_0^1x^{n-1}\operatorname d\!x=\frac1n .

Start by doing partial fractions, and then convert it to an integral and simplify.

∑ n = 1 ∞ n ( 3 n − 2 ) ( 3 n − 1 ) ( 3 n ) = 1 3 ∑ n = 1 ∞ ( 1 3 n − 2 − 1 3 n − 1 ) \displaystyle\sum_{n=1}^\infty\frac n{(3n-2)(3n-1)(3n)}=\frac13\sum_{n=1}^\infty\left(\frac1{3n-2}-\frac1{3n-1}\right) ∑ n = 1 ∞ n ( 3 n − 2 ) ( 3 n − 1 ) ( 3 n ) = 1 3 ∑ n = 1 ∞ ∫ 0 1 ( x 3 n − 3 − x 3 n − 2 ) d ⁡ ⁣ x \displaystyle\phantom{\sum_{n=1}^\infty\frac n{(3n-2)(3n-1)(3n)}}=\frac13\sum_{n=1}^\infty\int_0^1(x^{3n-3}-x^{3n-2})\operatorname d\!x
∑ n = 1 ∞ n ( 3 n − 2 ) ( 3 n − 1 ) ( 3 n ) = 1 3 ∫ 0 1 ∑ n = 1 ∞ ( x 3 n − 3 − x 3 n − 2 ) d ⁡ ⁣ x \displaystyle\phantom{\sum_{n=1}^\infty\frac n{(3n-2)(3n-1)(3n)}}=\frac13\int_0^1\sum_{n=1}^\infty(x^{3n-3}-x^{3n-2})\operatorname d\!x
∑ n = 1 ∞ n ( 3 n − 2 ) ( 3 n − 1 ) ( 3 n ) = 1 3 ∫ 0 1 ( 1 x 3 − 1 x 2 ) ∑ n = 1 ∞ x 3 n d ⁡ ⁣ x \displaystyle\phantom{\sum_{n=1}^\infty\frac n{(3n-2)(3n-1)(3n)}}=\frac13\int_0^1\left(\frac1{x^3}-\frac1{x^2}\right)\sum_{n=1}^\infty x^{3n}\operatorname d\!x
∑ n = 1 ∞ n ( 3 n − 2 ) ( 3 n − 1 ) ( 3 n ) = 1 3 ∫ 0 1 ( 1 x 3 − 1 x 2 ) x 3 1 − x 3 d ⁡ ⁣ x \displaystyle\phantom{\sum_{n=1}^\infty\frac n{(3n-2)(3n-1)(3n)}}=\frac13\int_0^1\left(\frac1{x^3}-\frac1{x^2}\right)\frac{x^3}{1-x^3}\operatorname d\!x
∑ n = 1 ∞ n ( 3 n − 2 ) ( 3 n − 1 ) ( 3 n ) = 1 3 ∫ 0 1 1 x 2 + x + 1 d ⁡ ⁣ x \displaystyle\phantom{\sum_{n=1}^\infty\frac n{(3n-2)(3n-1)(3n)}}=\frac13\int_0^1\frac1{x^2+x+1}\operatorname d\!x


The antiderivative happens to be 2 3 arctan ⁡ ( 2 x + 1 3 ) \frac2{\sqrt3}\arctan(\frac{2x+1}{\sqrt3}) .

∑ n = 1 ∞ n ( 3 n − 2 ) ( 3 n − 1 ) ( 3 n ) = π 9 3 \displaystyle\sum_{n=1}^\infty\frac n{(3n-2)(3n-1)(3n)}=\boxed{\displaystyle\frac\pi{9\sqrt3}}

19 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

This problem would be much harder if it were asked to find the exact form of S S . Wonderful!

@Pi Han Goh

Honestly , I was busy on Google Plus chatting with @Trevor Arashiro and @Sharky Kesa and others that I forgot about posting a solution here .

The method used is pretty obvious . This problem uses the same method too .

Don't worry , I'll post a solution if I say so . It's just that I honestly forgot , if you want you can cross-check with Trevor and Sharky .

A Former Brilliant Member - 6 years ago
Brock Brown
May 18, 2015

Python 2.7: 

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from fractions import Fraction as frac
from math import floor
a = 1
b = 1
S = 0
infinity = 2000
while a <= infinity:
    print a,b,b+1,b+2
    S += frac(a,b*(b+1)*(b+2))
    a += 1
    b += 3
print "Answer:", floor(10000*S)

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Kunal Gupta
May 16, 2015

The Title Says it ALL

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Please refrain from posting solution with no relevant working.

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