Question

If $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} [\frac{mn}{3^n}(\frac{1}{3^m}-\frac{n}{n\cdot3^m+m\cdot3^n})]=\frac{p}{q}$ (where p and q are relatively prime), then q – 3p is

Answer 1

5

Answer 2

Let the given sum be $S$.

$S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \left[\frac{mn}{3^n}\left(\frac{1}{3^m}-\frac{n}{n\cdot3^m+m\cdot3^n}\right)\right]$

Let the term inside the summation be $a_{m,n}$.

$a_{m,n} = \frac{mn}{3^n}\left(\frac{1}{3^m}-\frac{n}{n\cdot3^m+m\cdot3^n}\right)$

$a_{m,n} = \frac{mn}{3^n} \cdot \frac{1}{3^m} - \frac{mn}{3^n} \cdot \frac{n}{n\cdot3^m+m\cdot3^n}$

$a_{m,n} = \frac{mn}{3^{m+n}} - \frac{mn^2}{3^n(n\cdot3^m+m\cdot3^n)}$

The sum is $S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n}$.

Since the summation is over all positive integers $m$ and $n$, the sum is symmetric with respect to swapping $m$ and $n$.

Let's consider the term $a_{n,m}$ obtained by swapping $m$ and $n$ in $a_{m,n}$:

$a_{n,m} = \frac{nm}{3^m}\left(\frac{1}{3^n}-\frac{m}{m\cdot3^n+n\cdot3^m}\right)$

$a_{n,m} = \frac{nm}{3^m} \cdot \frac{1}{3^n} - \frac{nm}{3^m} \cdot \frac{m}{m\cdot3^n+n\cdot3^m}$

$a_{n,m} = \frac{mn}{3^{m+n}} - \frac{m^2n}{3^m(m\cdot3^n+n\cdot3^m)}$

The sum $S$ can also be written as $\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} a_{m,n}$. By swapping the summation indices, we get $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{n,m}$.

So, $S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{n,m}$.

Consider the sum of $a_{m,n}$ and $a_{n,m}$:

$a_{m,n} + a_{n,m} = \left(\frac{mn}{3^{m+n}} - \frac{mn^2}{3^n(n\cdot3^m+m\cdot3^n)}\right) + \left(\frac{mn}{3^{m+n}} - \frac{m^2n}{3^m(m\cdot3^n+n\cdot3^m)}\right)$

$a_{m,n} + a_{n,m} = \frac{2mn}{3^{m+n}} - \left(\frac{mn^2}{3^n(n\cdot3^m+m\cdot3^n)} + \frac{m^2n}{3^m(n\cdot3^m+m\cdot3^n)}\right)$

To combine the terms in the parenthesis, find a common denominator: $3^m 3^n (n\cdot3^m+m\cdot3^n) = 3^{m+n}(n\cdot3^m+m\cdot3^n)$.

$\frac{mn^2}{3^n(n\cdot3^m+m\cdot3^n)} + \frac{m^2n}{3^m(n\cdot3^m+m\cdot3^n)} = \frac{mn^2 \cdot 3^m + m^2n \cdot 3^n}{3^m 3^n (n\cdot3^m+m\cdot3^n)}$

$= \frac{mn(n \cdot 3^m + m \cdot 3^n)}{3^{m+n}(n\cdot3^m+m\cdot3^n)}$

$= \frac{mn}{3^{m+n}}$

So, $a_{m,n} + a_{n,m} = \frac{2mn}{3^{m+n}} - \frac{mn}{3^{m+n}} = \frac{mn}{3^{m+n}}$.

Now sum $a_{m,n} + a_{n,m}$ over all $m, n \ge 1$:

$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} (a_{m,n} + a_{n,m}) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{mn}{3^{m+n}}$

The left side is $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{m,n} + \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} a_{n,m} = S + S = 2S$.

The right side is $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \left(\frac{m}{3^m}\right)\left(\frac{n}{3^n}\right)$.

This is a product of two independent sums:

$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{mn}{3^{m+n}} = \left(\sum_{m=1}^{\infty} \frac{m}{3^m}\right) \left(\sum_{n=1}^{\infty} \frac{n}{3^n}\right)$.

We use the formula for the sum of an arithmetic-geometric series $\sum_{k=1}^{\infty} kx^k = \frac{x}{(1-x)^2}$ for $|x|<1$.

For $x = 1/3$, $\sum_{k=1}^{\infty} k\left(\frac{1}{3}\right)^k = \frac{1/3}{(1-1/3)^2} = \frac{1/3}{(2/3)^2} = \frac{1/3}{4/9} = \frac{1}{3} \cdot \frac{9}{4} = \frac{3}{4}$.

So, $\sum_{m=1}^{\infty} \frac{m}{3^m} = \frac{3}{4}$ and $\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}$.

The right side of the equation is $\left(\frac{3}{4}\right)\left(\frac{3}{4}\right) = \frac{9}{16}$.

So, $2S = \frac{9}{16}$.

$S = \frac{9}{32}$.

The problem states that the sum is $\frac{p}{q}$ where $p$ and $q$ are relatively prime.

We have $S = \frac{9}{32}$.

Here $p=9$ and $q=32$.

To check if they are relatively prime, find their prime factors.

$9 = 3^2$

$32 = 2^5$

Since they have no common prime factors, 9 and 32 are relatively prime.

The question asks for the value of $q - 3p$.

$q - 3p = 32 - 3(9) = 32 - 27 = 5$.