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kuing
Post time 2014-11-1 01:55
终于找到了连锁反应的路，看看有没有问题。
如楼主所计算的，我们有
\[a_n=2^{2^{n-1}+2n},\]
则
\begin{align*}
&\frac1{\sqrt2}\cdot\sqrt{a_1+\sqrt{a_2+\sqrt{a_3+\sqrt{a_4+\cdots +\sqrt{a_n}}}}} \\
={}&\sqrt{2^{-1}2^{2^0+2\times 1}+\sqrt{2^{-2}2^{2^1+2\times 2}+\sqrt{2^{-2^2}2^{2^2+2\times 3}+\sqrt{2^{-2^3}2^{2^3+2\times 4}+\cdots +\sqrt{2^{-2^{n-1}}2^{2^{n-1}+2n}}}}}} \\
={}&\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}}}}}},
\end{align*}
只需求上式的极限，然而，假如我们对上式最后一个根号内加上 $2^{n+1}+1$，则有
\begin{align*}
&\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}+2^{n+1}+1}}}}} \\
={}&\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{(2^n+1)^2}}}}} \\
={}&\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+2^n+1}}}} \\
={}&\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{(2^{n-1}+1)^2}}}} \\
={}&\cdots \\
={}&\sqrt{2^2+\sqrt{2^4+2^3+1}} \\
={}&\sqrt{2^2+\sqrt{(2^2+1)^2}} \\
={}&\sqrt{2^2+2^2+1} \\
={}&\sqrt{(2+1)^2} \\
={}&3,
\end{align*}
由此，通过不断分子有理化，我们有
\begin{align*}
&3-\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}}}}}} \\
={}&\frac{\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}+2^{n+1}+1}}}}-\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}}}}}}{3+\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}}}}}}} \\
={}&\frac{\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}+2^{n+1}+1}}}-\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}}}}}{\left( 3+\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}}}}}} \right)(\cdots )} \\
={}&\cdots \\
={}&\frac{2^{n+1}+1}{\left( 3+\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}}}}}} \right)(\cdots )\cdots (\cdots )},
\end{align*}
这里分母中的那些括号都是为有理化而乘上的式子，不需要具体写出来，只需知道它们显然都是大于 $3$ 的，那么显然分母比分子高阶，于是取极限即得
\[\lim_{n\to+\infty}3-\sqrt{2^2+\sqrt{2^4+\sqrt{2^6+\cdots +\sqrt{2^{2(n-1)}+\sqrt{2^{2n}}}}}}=0,\]
所以
\[\sqrt{a_1+\sqrt{a_2+\sqrt{a_3+\cdots}}}=3\sqrt2.\] |
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Tesla35
Post time 2014-10-31 22:45
