Sequences and series (2)

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^2n}{3^m(n\cdot 3^m + m\cdot3^n)}

Simplify the expression above.

If your answer comes in the form of a b \dfrac{a}{b} , where a a and b b are coprime positive integers, then enter a + b a+b as your answer.


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The answer is 41.

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2 solutions

Guilherme Niedu
May 9, 2017

S = m = 1 n = 1 m 2 n 3 m ( n 3 m + m 3 n ) \large \displaystyle S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{m^2 n}{3^m ( n \cdot 3^m + m \cdot 3^n ) }

S = 1 2 [ m = 1 n = 1 m 2 n 3 m ( n 3 m + m 3 n ) + m n 2 3 n ( m 3 n + n 3 m ) ] \large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{m^2 n}{3^m ( n \cdot 3^m + m \cdot 3^n ) } + \frac{mn^2}{3^n ( m \cdot 3^n + n \cdot 3^m ) } \right ]

S = 1 2 [ m = 1 n = 1 3 n m 2 n + 3 m m n 2 3 m + n ( n 3 m + m 3 n ) ] \large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{ 3^n \cdot m^2n + 3^m \cdot mn^2}{3^{m+n}( n \cdot 3^m + m \cdot 3^n )} \right ]

S = 1 2 [ m = 1 n = 1 m n ( n 3 m + m 3 n ) 3 m + n ( n 3 m + m 3 n ) ] \large \displaystyle S = \frac12 \left [ \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn ( n \cdot 3^m + m \cdot 3^n ) }{3^{m+n}( n \cdot 3^m + m \cdot 3^n )} \right ]

S = 1 2 m = 1 n = 1 m n 3 m + n \large \displaystyle S = \frac12 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{mn}{3^{m+n}}

S = 1 2 m = 1 m 3 m n = 1 n 3 n \large \displaystyle S = \frac12 \sum_{m=1}^{\infty} \frac{m}{3^m} \sum_{n=1}^{\infty} \frac{n}{3^n}

Note that n = 0 x n = 1 1 x \sum_{n=0}^{\infty} x^n = \frac{1}{1-x} , for x < 1 |x| < 1 . Differentiating w.r.t x x and multiplying by x x on both sides, one gets that n = 0 n x n = x ( 1 x ) 2 \sum_{n=0}^{\infty} n x^n = \frac{x}{(1-x)^2} .

So:

S = 1 2 [ 1 / 3 ( 1 1 / 3 ) 2 1 / 3 ( 1 1 / 3 ) 2 ] \large \displaystyle S = \frac12 \left [ \frac{1/3}{(1-1/3)^2} \cdot \frac{1/3}{(1-1/3)^2} \right ]

S = 9 32 \color{#20A900} \boxed{ \large \displaystyle S = \frac{9}{32} }

Then:

a = 9 , b = 32 , a + b = 41 \color{#3D99F6} \large \displaystyle a = 9, b = 32, \boxed{ \large \displaystyle a+b=41}

why this method works?

Hitesh Yadav - 9 months ago

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Because since m m and n n both cover the same interval, they're interchangeable in the sum.

Guilherme Niedu - 9 months ago

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Thanks for the reply.

Hitesh Yadav - 9 months ago

Replace 'm' with 'n' and 'n' with 'm' ,

Now add these two expressions,

Then we get terms which will be in A.G.P.,

So, by simplifying we get answer to be 9 / 32 {9}/{32} .

So a + b a+b = 41 41 .

It would be 9 / 32 9/32 . Please correct it.

Rahil Sehgal - 4 years, 1 month ago

Same approach, did it above :)

Guilherme Niedu - 4 years, 1 month ago

It would be nice if you provide an explanation as to why the method you mentioned works :)

Aditya Sky - 4 years, 1 month ago

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