Infinite or not?

For $Y=1,b=C=2$ , the denominator becomes $A-1$ , and $X$ can assume any positive integer value. Hence the given sum equals $\frac {1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...$

$=\dfrac 14\times \dfrac {1}{1-\frac 12}=\dfrac 12=\boxed {0.5}$ .

Lets say that A $\alpha$ B Means that when A is an integer then B is Integer, and when A is not integer then B is not, A Can equal B and not equal (got it?)

• $\boxed{\frac{A^x - 1}{A^y - 1} \alpha \frac{x}{y}}$ New math😂(if you want the prove i can write it)

So : $\dfrac{A^{(b^{(C^{X} - 1) }-1) }-1}{A^{(b^{(C^{Y} - 1) }-1) }-1}$ $\alpha$ $\dfrac{b^{(C^{X} - 1) }-1) }{(b^{(C^{Y} - 1) }-1) }$ $\alpha$ $\dfrac{C^{X} - 1}{C^{Y} - 1 }$ $\alpha$ $\dfrac{x}{y}$

$\dfrac{A^{(b^{(C^{X} - 1) }-1) }-1}{A^{(b^{(C^{Y} - 1) }-1) }-1}$ $\alpha$ $\dfrac{x}{y}$

Lets assuming that X=k×Y because this makes $\frac{X}{Y}$ = $\frac{k×Y}{Y}$ = k (k is integer 2,3,4....)

K $\ne$ 1 because that makes X=Y and we know that X $\ne$ Y

So : X+Y=(k+1)Y (k=2,3,4,5....) Changing (k+1) to n which they = 3,4,5,6... ( Only For make the expression easier)

X+Y=n×Y (Y can be 1,2,3,4,5,6....) (n=3,4,5,6...)

So - $\displaystyle\sum_{(X, Y) }$ $\frac{1}{2^{x+y}}$ = $\frac{1}{2^{1×3}}$ + $\frac{1}{2^{1×4}}$ + $\frac{1}{2^{1×5}}$ ..... + $\frac{1}{2^{2×3}}$ + $\frac{1}{2^{2×4}}$ + $\frac{1}{2^{2×5}}$ ...... + $\frac{1}{2^{3×3}}$ + $\frac{1}{2^{3×4}}$ + $\frac{1}{2^{3×5}}$ .........= $\displaystyle\sum_{Y=1}^{\infty}$ $\displaystyle\sum_{n=3}^{\infty} \frac{1}{2^{nY}}$

$\displaystyle\sum_{Y=1}^{\infty}$ $\displaystyle\sum_{n=3}^{\infty} \frac{1}{2^{nY}}$ = $\displaystyle\sum_{Y=1}^{\infty}$ $\displaystyle\sum_{n=3}^{\infty}$ $(\frac{1}{2^{Y} })^{n}$ = $\displaystyle\sum_{n=3}^{\infty} \frac{1}{2^n-1}$

THAT MEANS : $\displaystyle\sum_{(X, Y) }$ $\frac{1}{2^{x+y}}$ = $\displaystyle\sum_{n=3}^{\infty} \frac{1}{2^n-1}$ $\approx$ $\boxed{.27}$