I have two sequences $x_i$ and $y_i$ defined by their expressions : $$x_i-x_{i+1}=y_i-y_{i+1}=\sqrt{x_{i+1}y_{i+1}}$$ I have to prove that $xy(x-y)=0$. I tried this : I have $$x_i=x^{2^{i-1}}\prod_{j=0}^{j=i-2} (x^{2^j}+y^{2^j})^{-1}$$ And $$y_i=y^{2^{i-1}}\prod_{j=0}^{j=i-2} (x^{2^j}+y^{2^j})^{-1}$$ How to prove that $x=y$ ? I tried this : I suppose $x\neq{y}$ and I define the series. I see that $$\sqrt{x_iy_i}=y_{i-1}-y_i=x_{i-1}-x_i$$ So $$x_i-x_{i+1}=\sqrt{x_{i+1}y_{i+1}}$$ $$x_{i-1}-x_i=\sqrt{x_iy_i}$$ $$\vdots$$ $$x_1-x_2=x-x_2=\sqrt{x_2y_2}$$ Or $$\sum_{j=2}^{j=i+1}{(\sqrt{x_jy_j})}=x-x_2+x_2-x_3+\cdots+x_i-x_{i+1}=x-x_{i+1}$$ Its limit is $$\sum_{j=2}^{j=\infty}{(\sqrt{x_jy_j})}=\lim_{i\longrightarrow{\infty}}{(x-x_{i+1})}$$ If $x\geq{y}$ $$\sum_{j=2}^{j=\infty}{(\sqrt{x_jy_j})}=\lim_{i\longrightarrow{\infty}}{(x-x_{i+1})}=x-(x-y)=y$$ If $x\leq{y}$ $$\sum_{j=2}^{j=\infty}{(\sqrt{x_jy_j})}=\lim_{i\longrightarrow{\infty}}{(x-x_{i+1})}=x$$ I suppose $x\geq{y}$. Hence $$\sum_{j=2}^{j=i}{((-1)^j\sqrt{x_jy_j})}=x-x_2-(x_2-x_3)+(x_3-x_4)-\cdots+(-1)^i(x_{i-1}-x_i)$$ $$=x-2x_2+2x_3-\cdots+2(-1)^{i-1}x_{i-1}+(-1)^{i+1}x_i$$ $$=2\sum_{j=2}^{j=i-1}{((-1)^{j+1}x_j)}+x+(-1)^{i+1}x_i$$ $$=2\sum_{j=1}^{j=i}{((-1)^{j+1}x_j)}-x-(-1)^{i+1}x_i$$ $$=\sum_{j=2}^{j=i-1}{((-1)^{j+1}x_j)}+\sum_{j=1}^{j=i}{((-1)^{j+1}x_j)}$$ $$=2\sum_{j=2}^{j=i-1}{((-1)^{j+1}y_j)}+y+(-1)^{i+1}y_i$$ $$=2\sum_{j=1}^{j=i}{((-1)^{j+1}y_j)}-y-(-1)^{i+1}y_i$$ $$=\sum_{j=2}^{j=i-1}{((-1)^{j+1}y_j)}+\sum_{j=1}^{j=i}{((-1)^{j+1}y_j)}$$ Or $$2\sum_{j=1}^{j=i}{((-1)^{j+1}x_j)}=\sum_{j=2}^{j=i}{((-1)^j\sqrt{x_jy_j})}+x+(-1)^{i+1}x_i$$ And $$2\sum_{j=1}^{j=i}{((-1)^{j+1}y_j)}=\sum_{j=2}^{j=i}{((-1)^j\sqrt{x_jy_j})}+y+(-1)^{i+1}y_i$$ I do not know the limit of $(-1)^{i+1}x_i$, $$\sum_{j=1}^{j=\infty}{((-1)^{j}x_j)}$$ may diverge. But $$\sum_{j=2}^{j=\infty}{((-1)^j\sqrt{x_jy_j})}$$ is absolutely convergent. As $y_i$ tends to zero in the infinity, then $$\sum_{j=1}^{j=\infty}{((-1)^{j}y_j)}$$ converge. The limit of $$\sum_{j=2}^{j=i}{((-1)^j\sqrt{x_jy_j})}=2\sum_{j=1}^{j=i}{((-1)^{j+1}y_j)}-y-(-1)^{i+1}y_i$$ $$=2\sum_{j=1}^{j=i}{((-1)^{j+1}x_j)}-x-(-1)^{i+1}x_i$$ exists and the series are convergent. I will try to prove that $$\lim_{i\longrightarrow{\infty}}{(x_i)}=x-y=0$$ Let $$\sum_{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-\frac{k}{\sqrt{2m}}})}$$ $$=xe^{-\frac{1}{\sqrt{2m}}}-x_2e^{-\frac{2}{\sqrt{2m}}}+x_3e^{-\frac{3}{\sqrt{2m}}}-\cdots+(-1)^{2m+1}x_{2m}e^{-\frac{2m}{\sqrt{2m}}}$$ $$=xe^{-\frac{2}{\sqrt{2m}}}+x(e^{-\frac{1}{\sqrt{2m}}}-e^{-\frac{2}{\sqrt{2m}}})-x_2e^{-\frac{2}{\sqrt{2m}}}+x_3e^{-\frac{4}{\sqrt{2m}}}+x_3(e^{-\frac{3}{\sqrt{2m}}}-e^{-\frac{4}{\sqrt{2m}}})-x_4e^{-\frac{4}{\sqrt{2m}}}+\cdots-x_{2m}e^{-\frac{2m}{\sqrt{2m}}}$$ $$=xe^{-\frac{2}{\sqrt{2m}}}(e^{\frac{1}{\sqrt{2m}}}-1)+x_3e^{-\frac{4}{\sqrt{2m}}}(e^{\frac{1}{\sqrt{2m}}}-1)+\cdots+x_{2m-1}e^{-\frac{2m}{\sqrt{2m}}}(e^{\frac{1}{\sqrt{2m}}}-1)+$$ $$+(x-x_2)e^{-\frac{2}{\sqrt{2m}}}+(x_3-x_4)e^{-\frac{4}{\sqrt{2m}}}+\cdots+(x_{2m-1}-x_{2m})e^{-\frac{2m}{\sqrt{2m}}}$$ $$=(e^{\frac{1}{\sqrt{2m}}}-1)(xe^{-\frac{2}{\sqrt{2m}}}+x_3e^{-\frac{4}{\sqrt{2m}}}+\cdots+x_{2m-1}e^{-\sqrt{2m}})+(\sqrt{x_2y_2}e^{-\frac{2}{\sqrt{2m}}}+\sqrt{x_4y_4}e^{-\frac{4}{\sqrt{2m}}}+\cdots+\sqrt{x_{2m}y_{2m}}e^{-\frac{2m}{\sqrt{2m}}})$$ $$=(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})}+\sum_{k=1}^{k=m}{(\sqrt{x_{2k}y_{2k}}e^{-\frac{2k}{\sqrt{2m}}})}$$ Also $$\sum_{k=1}^{k=2m}{((-1)^{k+1}y_{k}e^{-\frac{k}{\sqrt{2m}}})}$$ $$=ye^{-\frac{1}{\sqrt{2m}}}-y_2e^{-\frac{2}{\sqrt{2m}}}+y_3e^{-\frac{3}{\sqrt{2m}}}-\cdots+(-1)^{2m+1}y_{2m}e^{-\frac{2m}{\sqrt{2m}}}$$ $$=(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})}+\sum_{k=1}^{k=m}{(\sqrt{x_{2k}y_{2k}}e^{-\frac{2k}{\sqrt{2m}}})}$$ But $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})}=S$$ And $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k+1}{\sqrt{2m}}})}<S<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k-1}{\sqrt{2m}}})}$$ $$(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{3}{\sqrt{2m}}}\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})}<S<(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{1}{\sqrt{2m}}}\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})}$$ Thus $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)y+(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=2}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=2}^{k=m}{(y_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{(S)}$$ And $$(e^{\frac{1}{\sqrt2m}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p}{\sqrt{2m}}})}=A$$ Or $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p+1}{\sqrt{2m}}})}<A<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}$$ Also $$(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{2}{\sqrt{2m}}}\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}<A<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}$$ Hence $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)y_{2p-1}e^{-\frac{p-1}{\sqrt{2m}}}+(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p+1}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p+1}^{k=m}{(y_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{(A)}=\lim_{m\longrightarrow{\infty}}{(S)}$$ For $$p+1=m\Rightarrow{\lim_{m\longrightarrow{\infty}}{(A)}=\lim_{m\longrightarrow{\infty}}{(S)}=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)y_{2m-1}e^{-\frac{m}{\sqrt{2m}}})}=0}$$ Consequently $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(y_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})})}=0$$ Also $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})}=S$$ Or $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k+1}{\sqrt{2m}}})}<S<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k-1}{\sqrt{2m}}})}$$ $$(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{3}{\sqrt{2m}}}\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})}<S<(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{1}{\sqrt{2m}}}\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})}$$ And $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)x+(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=2}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=2}^{k=m}{(x_{2k-1}e^{-\frac{2k-2}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{(S)}$$ Or $$(e^{\frac{1}{\sqrt2m}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p}{\sqrt{2m}}})}=A$$ And $$(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p+1}{\sqrt{2m}}})}<A<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}$$ Or $$(e^{\frac{1}{\sqrt{2m}}}-1)e^{-\frac{2}{\sqrt{2m}}}\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}<A<(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})}$$ Hence $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)x_{2p-1}e^{-\frac{p-1}{\sqrt{2m}}}+(e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p+1}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=p+1}^{k=m}{(x_{2k-1}e^{-\frac{2k-p-1}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{(A)}=\lim_{m\longrightarrow{\infty}}{(S)}$$ For $$p+1=m\Rightarrow{\lim_{m\longrightarrow{\infty}}{(A)}=\lim_{m\longrightarrow{\infty}}{(S)}=\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)x_{2m-1}e^{-\frac{m}{\sqrt{2m}}})}=0}$$ Consequently $$\lim_{m\longrightarrow{\infty}}{((e^{\frac{1}{\sqrt{2m}}}-1)\sum_{k=1}^{k=m}{(x_{2k-1}e^{-\frac{2k}{\sqrt{2m}}})})}=0$$ I deduce $$0<\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=2m}{((-1)^{k+1}x_{k}e^{-\frac{k}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=2m}{((-1)^{k+1}y_{k}e^{-\frac{k}{\sqrt{2m}}})})}$$ $$=\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=m}{(\sqrt{x_{2k}y_{2k}}e^{-\frac{2k}{\sqrt{2m}}})})}<\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=m}{(\sqrt{x_{2k}y_{2k}})})}$$ $$<\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=2m}{(\sqrt{x_{k}y_{k}})})}=y$$ Thus $$\lim_{m\longrightarrow{\infty}}{(\sum_{k=1}^{k=2m}{((-1)^{k+1}(x_k-y_k)e^{-\frac{k}{\sqrt{2m}}})})}=0$$ $$=\lim_{m\longrightarrow{\infty}}{((x-y)\sum_{k=1}^{k=2m}{(e^{-\frac{k}{\sqrt{2m}}})})}=\lim_{m\longrightarrow{\infty}}{((x-y)e^{-\frac{1}{\sqrt{2m}}}\frac{1-e^{-\sqrt{2m}}}{1+e^{-\frac{1}{\sqrt{2m}}}})}=\frac{x-y}{2}=0$$ And $$x-y=0$$ I recapitulate
$$x_i>x_{i+1},\quad{y_i>y_{i+1}}\Rightarrow{x=y}$$
$$x_i=x_{i+1}=x_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{xy=0}$$
$$y_i=y_{i+1}=y_{i+1}+\sqrt{x_{i+1}y_{i+1}}\Rightarrow{xy=0}$$ $$\Rightarrow{xy(x-y)=0}$$ Is this calculus correct ? Thank you.
