Can this be really done?

Calculus Level 5

∑ m = 1 ∞ ∑ n = 1 ∞ m 2 n 3 m ( n ⋅ 3 m + m ⋅ 3 n ) \large \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n \cdot 3^m + m \cdot 3^n) }

If the series above equals to α β \frac { \alpha }{ \beta } for coprime positive integers α , β \alpha, \beta , find β − α \beta - \alpha .

Try this set.


The answer is 23.

Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Aditya Kumar
May 17, 2015

∑ m = 1 ∞ ∑ n = 1 ∞ m 2 n 3 m ( n . 3 m + m . 3 n ) = 1 2 [ ∑ m = 1 ∞ ∑ n = 1 ∞ m 2 n 3 m ( n . 3 m + m . 3 n ) + ∑ m = 1 ∞ ∑ n = 1 ∞ n 2 m 3 n ( n . 3 m + m . 3 n ) ] = 1 2 [ ∑ m = 1 ∞ ∑ n = 1 ∞ m n 3 m + n ] = 1 2 [ ∑ m = 1 ∞ m 3 m ] 2 = 9 32 \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n.{ 3 }^{ m }+m.{ 3 }^{ n }) } } } =\quad \frac { 1 }{ 2 } \left[ \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n.{ 3 }^{ m }+m.{ 3 }^{ n }) } } } +\sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 }m }{ { 3 }^{ n }(n.{ 3 }^{ m }+m.{ 3 }^{ n }) } } } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 1 }{ 2 } \left[ \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { mn }{ { 3 }^{ m+n } } } } \right] \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 1 }{ 2 } { \left[ \sum _{ m=1 }^{ \infty }{ \frac { m }{ { 3 }^{ m } } } \right] }^{ 2 }\quad \quad \quad \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 9 }{ 32 } \quad \quad \quad

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I have tried limit test and have found series diverges :(

Ela Marinić-Kragić - 5 years, 11 months ago

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@Ela Marinić-Kragić I don't think so this is a limits question. It's a normal summation.

Aditya Kumar - 5 years, 10 months ago

Can you please elaborate your solution?

Kïñshük Sïñgh - 6 years ago

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What part of the solution do u want to be elaborated?

Aditya Kumar - 6 years ago

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Just curious to know how did u interchange m with n, and how do you know that the sum will be equal

Deep seth - 5 years, 9 months ago

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@Deep Seth – Since the sum of m and in is symmetric, we can always interchange them.

Aditya Kumar - 5 years, 9 months ago

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