Stunning Summation

Algebra Level 5

∑ n = 1 ∞ 8 n ( 3 n − 1 ) ! = A B e − C ( e D − E sin ⁡ ( π F + G ) ) \sum _{ n=1 }^{ \infty }{ \frac { { 8 }^{ n } }{ \left( 3n-1 \right) ! } } =\frac { A }{ B } { e }^{ -C }\left( { e }^{ D }-E\sin { \left( \frac { \pi }{ F } +\sqrt { G } \right) } \right)

If A , B , C , D , E , F A,B,C,D,E,F and G G are positive integers satisfying the equation above, find min ⁡ ( A + B + C + D + E + F + G ) \min(A+B+C+D+E+F+G ) .

Hint : Algebraic approach is easier than calculus approach.


The answer is 20.

Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Ariel Gershon
Feb 15, 2016

Replace 8 n 8^n with 2 3 n 2^{3n} , and this looks like the Taylor series of x e x xe^x with only one-third of the terms. My strategy is to find a function f ( x ) f(x) whose Taylor series is ∑ n = 1 ∞ x 3 n ( 3 n − 1 ) ! \sum_{n=1}^{\infty} \frac{x^{3n}}{(3n-1)!}

First, denote w = − 1 + − 3 2 w = \frac{-1+\sqrt{-3}}{2} , so that w 2 = − 1 − − 3 2 w^2 = \frac{-1-\sqrt{-3}}{2} and w 3 = 1 w^3 = 1 . Since the terms are nonzero periodically with a period of 3, I am going to make an educated guess that f ( x ) f(x) has the form p x e w x + q x e w 2 x + r x e x pxe^{wx} + qxe^{w^2x} + rxe^x . This function will have the following Taylor Series: ∑ k = 0 ∞ ( p w k + q w 2 k + r ) x k + 1 k ! \sum_{k=0}^{\infty} \dfrac{(pw^k+qw^{2k} + r)x^{k+1}}{k!}

Substituting k = 3 n , 3 n + 1 k = 3n, 3n+1 and 3 n − 1 3n-1 gives us the following system of equations: p + q + r = 0 w p + w 2 q + r = 0 w 2 p + w q + r = 1 \begin{array}{lr} p + q + r = 0 \\ wp + w^2q + r = 0 \\ w^2p + wq + r = 1 \end{array} Solving this gives p = w 3 , q = w 2 3 , r = 1 3 p = \frac{w}{3}, q = \frac{w^2}{3}, r = \frac{1}{3} . Therefore, f ( x ) = x 3 ( w e w x + w 2 e w 2 x + e x ) f(x) = \frac{x}{3}\left(we^{wx} + w^2e^{w^2x} + e^x\right)

Therefore, f ( 2 ) = ∑ n = 0 ∞ 2 3 n ( 3 n − 1 ) ! = 2 3 ( w e 2 w + w 2 e 2 w 2 + e 2 ) f(2) = \sum_{n=0}^{\infty} \frac{2^{3n}}{(3n-1)!} = \frac{2}{3}\left(we^{2w} + w^2e^{2w^2} + e^2\right) = 2 3 ( e 2 π 3 e − 1 + 3 i + e − 2 π 3 e − 1 − 3 i + e 2 ) = \frac{2}{3}\left(e^{\frac{2\pi}{3}}e^{-1+\sqrt{3}i} + e^{\frac{-2\pi}{3}}e^{-1-\sqrt{3}i} + e^2\right) = 2 3 e ( e ( 2 π 3 + 3 ) i + e ( − 2 π 3 − 3 ) i + e 3 ) = \frac{2}{3e}\left(e^{\left(\frac{2\pi}{3}+\sqrt{3}\right)i}+e^{\left(\frac{-2\pi}{3}-\sqrt{3}\right)i}+e^3\right) = 2 3 e [ cos ⁡ ( 2 π 3 + 3 ) + i sin ⁡ ( 2 π 3 + 3 ) + cos ⁡ ( − 2 π 3 − 3 ) + i sin ⁡ ( − 2 π 3 − 3 ) + e 3 ] = \frac{2}{3e} \left[\cos\left(\frac{2\pi}{3}+\sqrt{3}\right)+i\sin\left(\frac{2\pi}{3}+\sqrt{3}\right)+\cos\left(\frac{-2\pi}{3}-\sqrt{3}\right)+i\sin\left(\frac{-2\pi}{3}-\sqrt{3}\right) + e^3\right] = 2 3 e [ e 3 + 2 cos ⁡ ( 2 π 3 + 3 ) ] = \frac{2}{3e} \left[e^3 + 2\cos\left(\frac{2\pi}{3}+\sqrt{3}\right)\right] = 2 3 e [ e 3 − 2 sin ⁡ ( π 6 + 3 ) ] = \frac{2}{3e} \left[e^3 - 2\sin\left(\frac{\pi}{6}+\sqrt{3}\right)\right]

Hence, the answer is 2 + 3 + 1 + 3 + 2 + 6 + 3 = 20 2+3+1+3+2+6+3 = \boxed{20} .

6 Helpful 0 Interesting 1 Brilliant 0 Confused

Perfect one! Nice solution!

Aditya Kumar - 5 years, 4 months ago

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Could you post the calculus approach as well?

Sahil Jain - 3 years, 7 months ago

Hi Ariel, are you on Slack ? It's a social platform where most of the active users here (even the staffs) are there, and we talk about nothing except for math and science? Care to join us? It's free! Plus, there's a private channel where we discuss and share about difficult interesting series/integrals, you will definitely love it there!

Pi Han Goh - 5 years, 2 months ago

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No, I'm not on Slack. Sure, sounds interesting!

Ariel Gershon - 5 years, 2 months ago

I recently saw a solution to a similar problem, which gave me an idea for a better solution to this one:

∑ n = 1 ∞ 2 3 n ( 3 n − 1 ) ! = ∑ k = 0 ∞ 2 k + 1 k ! ( 1 3 + 2 3 cos ⁡ ( 2 π 3 ( k + 1 ) ) ) \sum_{n=1}^{\infty} \frac{2^{3n}}{(3n-1)!} = \sum_{k=0}^{\infty} \frac{2^{k+1}}{k!} \left(\frac{1}{3} + \frac{2}{3} \cos\left(\frac{2\pi}{3}(k+1)\right)\right) = 2 3 ∑ k = 0 ∞ 2 k k ! + 4 3 ∑ k = 0 ∞ 2 k k ! R e [ e 2 π 3 ( k + 1 ) i ] = \frac{2}{3} \sum_{k=0}^{\infty} \frac{2^k}{k!} + \frac{4}{3} \sum_{k=0}^{\infty} \frac{2^k}{k!} Re\left[e^{\frac{2\pi}{3}(k+1)i}\right] = 2 e 2 3 + R e [ 4 3 e 2 π i 3 ∑ k = 0 ∞ ( 2 e 2 π i 3 ) k k ! ] = \frac{2e^2}{3} + Re\left[\frac{4}{3}e^{\frac{2\pi i}{3}}\sum_{k=0}^{\infty} \frac{\left(2e^{\frac{2\pi i}{3}}\right)^k}{k!}\right] = 2 e 2 3 + R e [ 4 3 e 2 π i 3 e − 1 + 3 i ] = \frac{2e^2}{3} + Re\left[\frac{4}{3}e^{\frac{2\pi i}{3}} e^{-1+\sqrt{3}i}\right] = 2 e 2 3 + R e [ 4 3 e e ( 2 π 3 + 3 ) i ] = \frac{2e^2}{3} + Re\left[\frac{4}{3e}e^{\left(\frac{2\pi}{3}+\sqrt{3}\right)i}\right] = 2 e 2 3 + 4 3 e cos ⁡ ( 2 π 3 + 3 ) = \frac{2e^2}{3} + \frac{4}{3e} \cos{\left(\frac{2\pi}{3}+\sqrt{3}\right)} = 2 3 e [ e 3 − 2 sin ⁡ ( π 6 + 3 ) ] = \frac{2}{3e}\left[e^3 - 2\sin\left(\frac{\pi}{6}+\sqrt{3}\right) \right]

Ariel Gershon - 5 years, 2 months ago

This is way better than the previous one.

Gogole PI Mukherjee - 2 years, 3 months ago

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