if $\sum_{n=1}^{\infty}|x^n-y^n|<\infty$ then $|y|,|x|<1$ I want to prove for $x\neq y$ that if $\sum_{n=1}^{\infty}|x^n-y^n|<\infty$  then $|y|,|x|<1$. To do so, I started a proof by contradiction and arguing whether $x<y$ or not and whether these are negative or not ... 
But I guess there should be a simple, starightforward proof. I appreciate your help.
 A: The problem is not true if $x = y$. So I guess you meant to say $x, y$ are different. 
Hint: Using the reverse triangle inequality, you can deduce that $\sum_n ||x|^n - |y|^n|<\infty.$ Also, W.L.O.G, you can assumes $|x|<|y|$, so you can get rid of the absolue value sign. Now you can look at the partial sums and should be easy from there. 
Added:. The lesser known triangle inequality is this: $||x|-|y||\leq|x-y|$. If you use this, then the sum $\sum_n ||x|^n - |y|^n|$ is absolutely convergent. Hence, the sum $\sum_n(|x|^n-|y^n|)$ is convergent. Depending on whether $|x|<|y|$, each summand is either positive or negative. So you can assume that $|x|>|y|$. Then, look at the partial sum $\sum_{n=1}^{n=N} (|x|^n - |y|^n)$ is a convergent sequence. But this is equivalent to saying that each of the sum $\sum |x|^n$ and $\sum |y|^n$ is convergent. But that is true only when $|x|, |y|<1.$
A: Counter example, $x = 2, y = 2$, so I think you mean $x \neq y$.
Then you can write $y = x + \epsilon$, and take it from there, given that terms like $n \epsilon x^{n-1}$ don't converge.
A: For $x\ne y$ it is obvious if $x=0$ or $y=0.$ 
For $x\ne y$ and $x\ne 0\ne y$:
(i). If $x<0<y$ or $y<0<x$ then for ODD $n\in N$ we have $|x^n-y^n|=|x|^n+|y|^n$ which cannot converge to $0$ unless $|x|<1>|y|.$
(ii)(a). For $x, y$ both positive,or both negative,with $x \ne y,$ let $x'=\max (x,y)$ and $y'= \min (x,y).$  We have $\{x,y\}=\{x',y'\}$.Now let $1+d=x'/y'.$ We have $d>0$  and for $n\in N$ we have $$|x^n-y^n|=|x'^n-y'^n|=((1+d)^n-1)\cdot |y'|^n\geq n d |y'|^n$$ and this must tend to $0$ as $n\to \infty$, so $|y'|<1.$ 
(ii)(b). Now we must also have $|x'|<1$. Because  for all sufficiently large $n$, we  have $|y'|^n<1/2.$ So if $|x'|\geq 1,$ then for all but finitely many $n$ we would have  $$|x'^n-y'^n|\geq |x'|^n-|y'|^n\geq 1-|y'|^n>1/2.$$
(ii)(c). Since $\{x,y\}=\{x',y'\}$  and $|x'|<1>|y'|,$ we are done. 
