Double summation!

Algebra Level 4

∑ m = 0 ∞ ∑ n = 0 ∞ m + n + m n 2 m ( 2 m + 2 n ) = ? \large \sum_{m=0}^{\infty}\displaystyle\sum_{n=0}^{\infty} \dfrac{m+n+mn}{2^m(2^m+2^n)}= \ ?


The answer is 6.0000.

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Pranjal Jain by Pranjal Jain , 23, India

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2 solutions

U Z
Jan 9, 2015

Considering m and n as identical and independent

S = ∑ m = 0 ∞ ∑ n = 0 ∞ m + n + m n 2 m ( 2 m + 2 n ) S = \displaystyle\sum_{m=0}^{\infty}\displaystyle\sum_{n=0}^{\infty} \dfrac{m+n+mn}{2^m(2^m+2^n)}

2 S = ∑ m = 0 ∞ ∑ n = 0 ∞ m + n + m n 2 m + 2 n × [ 1 2 m + 1 2 n ] 2S= \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \dfrac{m+n+mn}{2^m+2^n} \times \Bigg[ \frac{1}{2^m} + \frac{1}{2^n} \Bigg]

= ∑ m = 0 ∞ ∑ n = 0 ∞ ( m + n + m n ) × 1 2 n × 1 2 m = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} (m+n+mn) \times \frac{1}{2^n} \times \frac{1}{2^m}

= ∑ m = 0 ∞ m ( 1 2 ) m ∑ n = 0 ∞ ( 1 2 ) n + ∑ m = 0 ∞ ( 1 2 ) m ∑ n = 0 ∞ n ( 1 2 ) n + ∑ m = 0 ∞ m ( 1 2 ) m ∑ n = 0 ∞ n ( 1 2 ) n =\sum_{m=0}^{\infty} m\Big(\frac{1}{2} \Big)^{m} \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^{n} +\sum_{m=0}^{\infty} \Big(\frac{1}{2}\Big)^{m} \sum_{n=0}^{\infty} n \Big(\frac{1}{2} \Big)^{n} + \sum_{m=0}^{\infty} m \Big( \frac{1}{2}\Big)^{m} \sum_{n=0}^{\infty} n \Big(\frac{1}{2} \Big)^{n}

Thus

2 S = 4 + 4 + 2.2 = 12 2S=4+4+2.2=12

S = 6 \boxed{S=6}

56 Helpful 2 Interesting 2 Brilliant 0 Confused

Hey Pranjal from where did You get this Problem ? I'am amazed That My Math Teacher Takes This question as Class illustration recently . And Now I think That She Also Uses Brilliant. ! ⌣ ¨ \ddot\smile

Deepanshu Gupta - 6 years, 5 months ago

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You should ask your teacher!
If she does, good for you :)
If she doesn't, introduce her to the wonder that is Brilliant! :)

Calvin Lin Staff - 6 years, 4 months ago

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Yes , I will surely asked him in her Next Lecture ! And Actually My Previous Math Teacher @Vagish Jha Sir is already on Brilliant, I respect him Very much But Unfortunately He Had Left our Institute , and actually he Taught Me in my 12th Class , and suggested me To use Brilliant , From That Day I used Brilliant for regularly bases ! ⌣ ¨ \ddot\smile

Deepanshu Gupta - 6 years, 4 months ago

So far I have recommended Brilliant to 500 people.

Vijay Simha - 5 years, 10 months ago

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@Vijay Simha – That's really exciting! I hope they loved it as much as you do :)

Calvin Lin Staff - 5 years, 10 months ago

Can you explain me the Second last line

Department 8 - 5 years, 9 months ago

Great question and great solution. I found the result by brute force.

Shib Shankar Sikder - 5 years, 10 months ago

How did you get the second line?

Jessica Wang - 5 years, 8 months ago

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Which one?

2 S = ∑ m = 0 ∞ ∑ n = 0 ∞ m + n + m n 2 m + 2 n ∗ ( 1 2 m + 1 2 n ) 2S = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{m+n+mn}{2^m +2^n} * (\frac{1}{2^m} + \frac{1}{2^n}) is just distributive law,

∑ m = 0 ∞ ∑ n = 0 ∞ m + n + m n 2 m + 2 n ∗ 1 2 m + 1 2 n \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{m+n+mn}{2^m +2^n} * \frac{1}{2^m} + \frac{1}{2^n}

= ∑ m = 0 ∞ ∑ n = 0 ∞ ( m + n + m n ) ∗ 1 2 m ∗ 2 n = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} (m+n+mn) * \frac{1}{2^m * 2^n} also looks like distributive law to me: encapsulating 1 2 m ∗ 2 n \frac{1}{2^m * 2^n} from above's equation. Both sides can be multiplied by 2 m ∗ 2 n 2^m * 2^n , then it is obvious that this step is true. Deriving those steps might just require a keen eye or really lots of practice..

Alisa Meier - 5 years, 8 months ago

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@Alisa Meier No I mean, how do you go from S = ∑ m = 0 ∞ ∑ n = 0 ∞ m + n + m n 2 m ( 2 m + 2 n ) S = \displaystyle\sum_{m=0}^{\infty}\displaystyle\sum_{n=0}^{\infty} \dfrac{m+n+mn}{2^m(2^m+2^n)} to 2 S = ∑ m = 0 ∞ ∑ n = 0 ∞ m + n + m n 2 m + 2 n × [ 1 2 m + 1 2 n ] 2S= \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \dfrac{m+n+mn}{2^m+2^n} \times \Bigg[ \frac{1}{2^m} + \frac{1}{2^n} \Bigg] ?

Jessica Wang - 5 years, 8 months ago

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@Jessica Wang – Oh, it is even easier. Symmetry. Just compare the terms and you see they are equal:

S = ∑ m = 0 ∞ ∑ n = 0 ∞ m + n + m n 2 m ( 2 m + 2 n ) S = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{m+n+mn}{2^m (2^m +2^n)} = ∑ n = 0 ∞ ∑ m = 0 ∞ m + n + m n 2 m ( 2 m + 2 n ) \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{m+n+mn}{2^m (2^m +2^n)}

S = ∑ m = 0 ∞ ∑ n = 0 ∞ m + n + m n 2 n ( 2 m + 2 n ) S = \sum_{m=0}^{\infty} \sum_{n=0}^{\infty} \frac{m+n+mn}{ 2^n (2^m +2^n)} = ∑ n = 0 ∞ ∑ m = 0 ∞ m + n + m n 2 n ( 2 m + 2 n ) \sum_{n=0}^{\infty} \sum_{m=0}^{\infty} \frac{m+n+mn}{2^n (2^m +2^n)}

For further reading: "Riemann Rearrangement Theorem"

Related: "Riemann series theorem"

Alisa Meier - 5 years, 8 months ago

Loved your solution !! Thanks !!

Akshat Sharda - 5 years, 3 months ago

The symmetrizing of n and m is readily seen when you type all the sum in a sieve and analize it as an infinite matrix; this method has always been of my predilection since i first thought of it, the key is the identity that makes the colapsing of the sum in the denominator to a power of 2 and so you can add it because the numerators form a symmetric matrix and use some nice identities on the sums of powers.

Héctor Andrés Parra Vega - 5 years, 10 months ago

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What other nice identities do you know involving this idea? Care to share?

Calvin Lin Staff - 5 years, 10 months ago
Vinay Agarwal
Jan 15, 2015

I have used the interactive phase of Java to evaluate the sum. That is I have used net beans IDE for the purpose. And the code used is as follows: 

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int i,j;
double sum=0;
for(i=0;i<=100000;i++)    // "i" in the code can be treated as "n"
{
//"j" in the code can be treated as "m"
for(j=0;j<=100000;j++)
{
double term=(i+j+i*j)/(Math.pow(2,i)*(Math.pow(2,i)+Math.pow(2,j)));
sum+=term;
}
}
System.out.println(""+sum);

As the terms after large values of "n" and "m" becomes too insignificant to affect the answer thus we can run the loop up to 100000 to have a fair idea of the answer.

5 Helpful 1 Interesting 0 Brilliant 0 Confused

Can you post the code? It would be appreciated.

Pranjal Jain - 6 years, 4 months ago

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I have used the interactive phase of Java to evaluate the sum. That is I have used net beans IDE for the purpose. And the code used is as follows: 

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
 
int i,j;
double sum=0;
for(i=0;i<=100000;i++)    // "i" in the code can be treated as "n"
{
//"j" in the code can be treated as "m"
for(j=0;j<=100000;j++)
{
double term=(i+j+i*j)/(Math.pow(2,i)*(Math.pow(2,i)+Math.pow(2,j)));
sum+=term;
}
}
System.out.println(""+sum);

As the terms after large values of "n" and "m" becomes too insignificant to affect the answer thus we can run the loop up to 100000 to have a fair idea of the answer.

Vinay Agarwal - 6 years, 4 months ago

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Vinay, please refrain from typing in all capital letters, as that is taken to mean that you are shouting and is generally considered disrespectful on the internet.

Calvin Lin Staff - 6 years, 4 months ago

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