# Note
>Relations between pure and applied mathematicians are based on trust and understanding. Namely, pure mathematicians do not trust applied mathematicians, and applied mathematicians do not understand pure mathematicians. -- Albert Einstein
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**1.** (5 points) Given $a_1>g_1>0$, we define $a_{n+1}:=\frac{1}{2}(a_n+g_n)$ and $g_{n+1}:=\sqrt{a_n g_n}$ for every $n\in\mathbb{N}$. Show that both $\lim\limits_{n\to\infty}{a_n}$ and $\lim\limits_{n\to\infty}{g_n}$ exists and are equal.
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By the AM--GM inequality, $a_n\ge g_n$ for every $n\in\mathbb{N}$. Since for every $n\in\mathbb{N}$ we have
$$a_{n+1}=\frac{1}{2}(a_n+g_n)\le\frac{1}{2}(a_n+a_n)=a_n$$ and
$$g_{n+1}=\sqrt{a_n g_n}\ge\sqrt{g_n g_n}=g_n,$$ $(a_n)$ is decreasing and $(g_n)$ is increasing.
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