An algebra problem by A Former Brilliant Member

Algebra Level 4

∏ n = 1 ∞ ( 3 n ) 1 / 3 n = 3 1 / 3 × 9 1 / 9 × 2 7 1 / 27 × ⋯ \large \prod_{n=1}^\infty (3^n)^{1/3^n} =3^{1/3} \times 9^{1/9} \times 27^{1/27} \times \cdots

The product above can be expressed as a b / c a^{b/c} , where a a is a prime number , and b , c b,c are coprime positive integers . Find a + b + c a + b +c .


The answer is 10.

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A Former Brilliant Member by A Former Brilliant Member

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3 solutions

Swagat Panda
Aug 13, 2016

3 1 / 3 × 3 2 / 9 × 3 3 / 27 ⋯ = E Let 3 S = E ⇒ S = 1 3 + 2 9 + 3 27 + ⋯ 3 S = 1 + 2 3 + 3 9 + ⋯ 3 S − S = 1 + 1 3 + 1 9 + 1 27 ⋯ 2 S = 1 1 − 1 3 = 3 2 S = 3 4 ⇒ E = 3 3 / 4 A + B + C = 3 + 3 + 4 = 10 {3^{1/3}\times3^{2/9}\times3^{3/27}\cdots=E \\ \text{Let }3^{S}=E \\ \Rightarrow S=\dfrac13+\dfrac29+\dfrac3{27}+\cdots \\ 3S=1+\dfrac23+\dfrac39+\cdots \\ 3S-S=1+\dfrac13+\dfrac19+\dfrac1{27}\cdots \\ 2S=\dfrac{1}{1-\dfrac13}=\dfrac32 \\ S=\dfrac34 \Rightarrow E={3}^{3/4} \\\boxed{ A+B+C=3+3+4=10}}

9 Helpful 0 Interesting 0 Brilliant 0 Confused

in the 5th row, it should be 3 S − S 3S-S isn't it?

Muhammad Altair - 4 years, 10 months ago

Yes, it was a typo error. Thanks for correcting me. :)

Swagat Panda - 4 years, 10 months ago
Chew-Seong Cheong
Aug 13, 2016

P = ∏ n = 1 ∞ ( 3 n ) 1 3 n log ⁡ P = log ⁡ ∏ n = 1 ∞ ( 3 n 3 n ) = ∑ n = 1 ∞ log ⁡ ( 3 n 3 n ) = log ⁡ 3 ∑ n = 1 ∞ n 3 n = 1 3 log ⁡ 3 ∑ n = 1 ∞ n 3 n − 1 Let x = 1 3 = 1 3 log ⁡ 3 ∑ n = 1 ∞ n x n − 1 = 1 3 log ⁡ 3 ∑ n = 1 ∞ d d x ( x n ) = 1 3 log ⁡ 3 d d x ∑ n = 1 ∞ x n = 1 3 log ⁡ 3 d d x ( x 1 − x ) = 1 3 log ⁡ 3 ( 1 1 − x + x ( 1 − x ) 2 ) = 1 3 log ⁡ 3 ( 1 ( 1 − x ) 2 ) Put back x = 1 3 = 1 3 log ⁡ 3 ( 1 ( 1 − 1 3 ) 2 ) = 1 3 log ⁡ 3 ( 9 4 ) log ⁡ P = 3 4 log ⁡ 3 ⟹ P = 3 3 / 4 \begin{aligned} P & = \prod_{n=1}^\infty \left(3^n\right)^\frac 1{3^n} \\ \log P & = \log \prod_{n=1}^\infty \left(3^\frac n{3^n}\right) = \sum _{n=1}^\infty \log \left(3^\frac n{3^n}\right) = \log 3 \sum _{n=1}^\infty \frac n{3^n} \\ & = \frac 1{\color{#D61F06}{3}} \log 3 \sum _{n=1}^\infty \frac n {\color{#3D99F6}{3}^{n\color{#D61F06}{-1}}} \quad \quad \small \color{#3D99F6}{\text{Let }x = \frac 13} \\ & = \frac 13 \log 3 \sum _{n=1}^\infty nx^{n-1} = \frac 13 \log 3 \sum _{n=1}^\infty \frac d{dx}(x^n) = \frac 13 \log 3 \frac d{dx} \sum _{n=1}^\infty x^n \\ & = \frac 13 \log 3 \frac d{dx} \left( \frac x{1-x} \right) = \frac 13 \log 3 \left( \frac 1{1-x} + \frac x{(1-x)^2} \right) \\ & = \frac 13 \log 3 \left( \frac 1{(1-\color{#3D99F6}{x})^2} \right) \quad \quad \small \color{#3D99F6}{\text{Put back }x = \frac 13} \\ & = \frac 13 \log 3 \left( \frac 1{\left(1-\color{#3D99F6}{\frac 13}\right)^2} \right) = \frac 13 \log 3 \left( \frac 94 \right) \\ \log P & = \frac 34 \log 3 \\ \implies P & = 3^{3/4} \end{aligned}

⟹ a + b + c = 3 + 3 + 4 = 10 \implies a+b+c = 3+3+4 = \boxed{10}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

a=3 b= 3 c= 4 feel free to ask anything

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Please tell me what is wrong in this solution... 3 1 / 3 × 3 2 / 9 × 3 3 / 27 ⋯ = E 3 S = E ⇒ S = 1 3 + 2 9 + 3 27 + ⋯ 3 S = 1 + 2 3 + 3 9 + ⋯ 3 S − S = 1 + 1 3 + 1 9 + 1 27 ⋯ 2 S = 1 1 − 1 3 = 3 2 S = 3 4 ⇒ E = 3 3 4 A + B + C = 3 + 3 + 4 = 10 3^{1/3}\times3^{2/9}\times3^{3/27}\cdots=E \\3^{S}=E \\ \Rightarrow S=\dfrac13+\dfrac29+\dfrac3{27}+\cdots \\3S=1+\dfrac23+\dfrac39+\cdots \\3S-S=1+\dfrac13+\dfrac19+\dfrac1{27}\cdots \\2S=\dfrac{1}{1-\dfrac13}=\dfrac32 \\S=\dfrac34 \Rightarrow E=3^\frac34 \\ A+B+C=3+3+4=10

Swagat Panda - 4 years, 10 months ago

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i have edited the prob , your solution is right, i forgot the real problem as i did it tomorrow , sorry for the inconvience. i will name you as a solver if possible

A Former Brilliant Member - 4 years, 10 months ago

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Thanks, I tried the answer 10 but it showed wrong, I checked my answer several times, but couldn't find any mistake, so I decided to post it, coz cross checking helps. Now I am a bit relieved.

Swagat Panda - 4 years, 10 months ago

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