A set of rational numbers

If we write $N'=\left \{ \frac{1}{2^1}, \frac{1}{2^2}, \ldots, \frac{1}{2^x} \right \}$ , the total of all elements of $Q$ is simply the product of the sum of all terms in $M$ and the sum of all terms in $N'$ , which is

$\frac{x(x+1)}{2} \cdot \left( 1- \frac{1}{2^x} \right)$

There are $x^2$ elements in $Q$ ; so the arithmetic mean is

$\overline{Q}=\frac{x(x+1)}{2x^2} \cdot \left( 1- \frac{1}{2^x} \right) = \frac{1+x^{-2}}{2} \cdot \left( 1- \frac{1}{2^x} \right)$

This clearly tends to $\boxed{\frac{1}{2}}$ as $x$ tends to infinity.

Extending $N$ to include $2^0$ changes the sum of $N'$ to $\left( 2- \frac{1}{2^x} \right)$ ; the new limit for $\overline{Q}$ would be $1$ .

$\bar { Q } =\frac { sumOfAllElementsOfQ }{ numberOfElementsOfQ }$

$M$ has $x$ elements and $N$ has $x$ elements as well. Thus, for every demominator we have x numerators,there are $x$ demoninators so there are ${ x }^{ 2 }$ elements in $Q$ .

To get the sum of all elements of $Q$ , lets focus to a fixed denominator. The sum of all fractions for a fixed denominator is:

$\frac { 1 }{ n } +\frac { 2 }{ n } +\frac { 3 }{ n } +\cdots +\frac { x }{ n } =\frac { 1+2+3+\cdots +x }{ n } =\frac { \sum _{ i=1 }^{ x }{ i } }{ n }$

The formula for the powers of 2 is ${ 2 }^{ 1 }+{ 2 }^{ 2 }+{ 2 }^{ 3 }+\cdots +{ 2 }^{ x }=\sum _{ n=1 }^{ x }{ \left\{ { 2 }^{ n } \right\} }$ . Thus, we get:

$\sum _{ n=1 }^{ x }{ \left\{ \frac { \sum _{ i=1 }^{ x }{ i } }{ { 2 }^{ n } } \right\} = } \sum _{ i=1 }^{ x }{ i } \cdot \sum _{ n=1 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ n } } \right\} } =\frac { x\cdot \left( x+1 \right) }{ 2 } \cdot \sum _{ n=1 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ n } } \right\} }$

If we combine everything together, we get:

$\bar { Q } =\frac { \frac { x\cdot \left( x+1 \right) }{ 2 } \cdot \sum _{ n=1 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ x } } \right\} } }{ { x }^{ 2 } }$

$\bar { Q } =\frac { x\cdot \left( x+1 \right) \cdot \sum _{ n=1 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ x } } \right\} } }{ 2\cdot { x }^{ 2 } }$

And now we have to calculate the limit:

$\lim _{ x\rightarrow \infty }{ \bar { Q } } =\lim _{ x\rightarrow \infty }{ \frac { x\cdot \left( x+1 \right) \cdot \sum _{ n=1 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ x } } \right\} } }{ 2\cdot { x }^{ 2 } } }$

We know, that $\sum _{ n=1 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ x } } \right\} }$ converges to $1$ (if $x$ grows up to infinity).

$\lim _{ x\rightarrow \infty }{ \bar { Q } } =\lim _{ x\rightarrow \infty }{ \frac { x\cdot \left( x+1 \right) \cdot 1 }{ 2\cdot { x }^{ 2 } } } =\lim _{ x\rightarrow \infty }{ \frac { x\cdot \left( x+1 \right) }{ 2\cdot { x }^{ 2 } } }=\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }+x }{ 2\cdot { x }^{ 2 } } }$

$\lim _{ x\rightarrow \infty }{ \bar { Q } } =\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }\cdot \left( 1+\frac { 1 }{ x } \right) }{ { x }^{ 2 }\cdot \left( 2 \right) } = } \lim _{ x\rightarrow \infty }{ \frac { \left( 1+\frac { 1 }{ x } \right) }{ 2 } }$

Deleting the zero sequence, leads to:

$\lim _{ x\rightarrow \infty }{ \bar { Q } } =\lim _{ x\rightarrow \infty }{ \frac { 1 }{ 2 } } =\boxed { \frac { 1 }{ 2 } }$

bonus question:

If we insert a single $1$ to $N$ , then the formula would change a little bit, because we have now: $\sum _{ n=0 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ n } } \right\} }$

Thus, we get:

$\bar { Q } =\frac { \frac { x\cdot \left( x+1 \right) }{ 2 } \cdot \sum _{ n=0 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ x } } \right\} } }{ { x }^{ 2 } }$

And now we have to calculate the limit as befor, but $\sum _{ n=0 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ n } } \right\} }$ converges to $2$ (if $x$ grows up to infinity).

$\lim _{ x\rightarrow \infty }{ \bar { Q } } =\lim _{ x\rightarrow \infty }{ \frac { x\cdot \left( x+1 \right) \cdot \sum _{ n=0 }^{ x }{ \left\{ \frac { 1 }{ { 2 }^{ x } } \right\} } }{ 2\cdot { x }^{ 2 } } }$

With the same conversions as befor:

$\lim _{ x\rightarrow \infty }{ \bar { Q } } =\lim _{ x\rightarrow \infty }{ \frac { x\cdot \left( x+1 \right) \cdot 2 }{ 2\cdot { x }^{ 2 } } } =\lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }+x }{ { x }^{ 2 } } = } \lim _{ x\rightarrow \infty }{ \frac { { x }^{ 2 }\cdot \left(1+ \frac { 1 }{ x } \right) }{ { x }^{ 2 } } = } \lim _{ x\rightarrow \infty }{ 1= } \boxed { 1 }$