Infinite exponent expression

Algebra Level 3

6 7 + 1 − 3 + 2 − 3 + 3 − 3 + 4 − 3 + 5 − 3 + ⋯ 1 − 3 + 3 − 3 + 5 − 3 + 7 − 3 + 9 − 3 + ⋯ = ? \dfrac{6}{7}\ + \dfrac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+\cdots }{1^{-3}+3^{-3}+5^{-3}+7^{-3}+9^{-3}+\cdots } = \, ?


The answer is 2.

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Azzam Labib by Azzam Labib , 19, Indonesia

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2 solutions

Chew-Seong Cheong
Jun 29, 2016

Relevant wiki: Algebraic Manipulation - Rearranging

X = 6 7 + 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + . . . 1 1 3 + 1 3 3 + 1 5 3 + 1 7 3 + 1 9 3 + . . . = 6 7 + 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + . . . ( 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + . . . ) − ( 1 2 3 + 1 4 3 + 1 6 3 + 1 8 3 + 1 1 0 3 + . . . ) = 6 7 + 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + . . . ( 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + . . . ) − 1 2 3 ( 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + . . . ) = 6 7 + 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + . . . 7 8 ( 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + . . . ) = 6 7 + 8 7 = 2 \begin{aligned} X & = \frac 67 + \frac {\frac 1{1^3} + \frac 1{2^3} +\frac 1{3^3} +\frac 1{4^3} +\frac 1{5^3} +...}{\frac 1{1^3} + \frac 1{3^3} +\frac 1{5^3} +\frac 1{7^3} +\frac 1{9^3} +...} \\ & = \frac 67 + \frac {\frac 1{1^3} + \frac 1{2^3} +\frac 1{3^3} +\frac 1{4^3} +\frac 1{5^3} +...}{\left(\frac 1{1^3} + \frac 1{2^3} +\frac 1{3^3} +\frac 1{4^3} +\frac 1{5^3} +...\right) - \left(\frac 1{2^3} + \frac 1{4^3} +\frac 1{6^3} +\frac 1{8^3} +\frac 1{10^3} +...\right)} \\ & = \frac 67 + \frac {\frac 1{1^3} + \frac 1{2^3} +\frac 1{3^3} +\frac 1{4^3} +\frac 1{5^3} +...}{\left(\frac 1{1^3} + \frac 1{2^3} +\frac 1{3^3} +\frac 1{4^3} +\frac 1{5^3} +...\right) - \frac 1{2^3}\left(\frac 1{1^3} + \frac 1{2^3} +\frac 1{3^3} +\frac 1{4^3} +\frac 1{5^3} +...\right)} \\ & = \frac 67 + \frac {{\frac 1{1^3} + \frac 1{2^3} +\frac 1{3^3} +\frac 1{4^3} +\frac 1{5^3} +...}}{\frac 78 \left({\frac 1{1^3} + \frac 1{2^3} +\frac 1{3^3} +\frac 1{4^3} +\frac 1{5^3} +...}\right)} \\ & = \frac 67 + \frac 87 = \boxed{2} \end{aligned}

33 Helpful 0 Interesting 0 Brilliant 0 Confused

Awesome! Thanks.

Siddhesh Deshpande - 4 years, 11 months ago

Great solution!

Bli Bla - 4 years, 11 months ago
Derek Modzelewski
Jun 30, 2016

Without doing any math, the solution is requested in decimal format, and does not specify number of digits so is almost definitely integral. The expression after 6/7 is certainly greater than 1, but not more than 2, leaving us with 2 as the only possible solution.

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Wow. Quite the unique solution we have here ;)

Manuel Kahayon - 4 years, 11 months ago

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