TO INFINITY AND BEYOND!

Algebra Level pending

I f ∑ n = 1 ∞ ( 1 2 ) n cos ⁡ 2 n π 3 c a n b e e x p r e s s e d i n t h e f o r m − a b w h e r e a a n d b a r e r e l a t i v e l y p r i m e i n t e g e r s , w h a t i s a + b ? If\quad \sum _{ n=1 }^{ \infty }{ { \left( \frac { 1 }{ 2 } \right) }^{ n }\cos { \frac { 2n\pi }{ 3 } } \quad } can\quad be\quad expressed\\ in\quad the\quad form\quad -\frac { a }{ b } \quad where\quad a\quad and\quad b\quad are\quad \\ relatively\quad prime\quad integers,\quad what\quad is\quad a+b?


The answer is 9.

Justin Tuazon by Justin Tuazon , 33, Philippines

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1 solution

Justin Tuazon
Dec 30, 2014

∑ n = 1 ∞ ( 1 2 ) n cos ⁡ 2 n π 3 = cos ⁡ 2 π 3 2 + cos ⁡ 4 π 3 2 2 + cos ⁡ 2 π 2 3 + cos ⁡ 8 π 3 2 4 + . . . = 1 2 ( − 1 2 ) + 1 2 2 ( − 1 2 ) + 1 2 3 + 1 2 4 ( − 1 2 ) + 1 2 5 ( − 1 2 ) + 1 2 6 + . . . = ( − 1 4 ) ( 1 + 1 2 3 + 1 2 6 + . . . ) = ( − 1 4 ) ( 1 1 − 1 8 ) = − 2 7 a = 2 , b = 7 ∴ a + b = 9 \sum _{ n=1 }^{ \infty }{ { \left( \frac { 1 }{ 2 } \right) }^{ n }\cos { \frac { 2n\pi }{ 3 } } } \\ =\frac { \cos { \frac { 2\pi }{ 3 } } }{ 2 } +\frac { \cos { \frac { 4\pi }{ 3 } } }{ { 2 }^{ 2 } } +\frac { \cos { 2\pi } }{ { 2 }^{ 3 } } +\frac { \cos { \frac { 8\pi }{ 3 } } }{ { 2 }^{ 4 } } +...\\ =\frac { 1 }{ 2 } \left( -\frac { 1 }{ 2 } \right) +\frac { 1 }{ { 2 }^{ 2 } } \left( -\frac { 1 }{ 2 } \right) +\frac { 1 }{ { 2 }^{ 3 } } +\frac { 1 }{ { 2 }^{ 4 } } \left( -\frac { 1 }{ 2 } \right) +\frac { 1 }{ { 2 }^{ 5 } } \left( -\frac { 1 }{ 2 } \right) +\frac { 1 }{ { 2 }^{ 6 } } +...\\ =\left( \frac { -1 }{ 4 } \right) \left( 1+\frac { 1 }{ { 2 }^{ 3 } } +\frac { 1 }{ { 2 }^{ 6 } } +... \right) \\ =\left( \frac { -1 }{ 4 } \right) \left( \frac { 1 }{ 1-\frac { 1 }{ 8 } } \right) =-\frac { 2 }{ 7 } \\ a=2,\quad b=7\\ \\ \therefore \quad a+b=9

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