[tex]\Large \begin{aligned}&\sqrt 2\cdot \sqrt[4]2\cdot \sqrt[8]2\cdot\ldots\cdot\sqrt[2^n]2=2^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}\cdot2^{\frac{1}{8}}\cdot\ldots\cdot2^{\frac{1}{2^n}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^n}}\end{aligned}[/tex]
W wykładniku mamy sumę ciągu geometrycznego gdzie:
[tex]a_1=\dfrac{1}{2}\\\\q=\dfrac{1}{2}[/tex]
[tex]S_n=\dfrac{a_1(1-q^n)}{1-q}\\\\S_n=\dfrac{\dfrac{1}{2}\left(1-\left(\dfrac{1}{2}\right)^n\right)}{1-\dfrac{1}{2}}=\dfrac{\dfrac{1}{2}\left(1-\left(\dfrac{1}{2}\right)^n\right)}{\dfrac{1}{2}}=1-\left(\dfrac{1}{2}\right)^n[/tex]
Zatem
[tex]\displaystyle\\\Large\begin{aligned}\lim_{n\to\infty}\sqrt 2\cdot \sqrt[4]2\cdot \sqrt[8]2\cdot\ldots\cdot\sqrt[2^n]2=\lim_{n\to\infty}2^{1-\left(\frac{1}{2}\right)^n}=2^{1-0}=2^1=2\end{aligned}[/tex]
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[tex]\Large \begin{aligned}&\sqrt 2\cdot \sqrt[4]2\cdot \sqrt[8]2\cdot\ldots\cdot\sqrt[2^n]2=2^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}\cdot2^{\frac{1}{8}}\cdot\ldots\cdot2^{\frac{1}{2^n}}=2^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^n}}\end{aligned}[/tex]
W wykładniku mamy sumę ciągu geometrycznego gdzie:
[tex]a_1=\dfrac{1}{2}\\\\q=\dfrac{1}{2}[/tex]
[tex]S_n=\dfrac{a_1(1-q^n)}{1-q}\\\\S_n=\dfrac{\dfrac{1}{2}\left(1-\left(\dfrac{1}{2}\right)^n\right)}{1-\dfrac{1}{2}}=\dfrac{\dfrac{1}{2}\left(1-\left(\dfrac{1}{2}\right)^n\right)}{\dfrac{1}{2}}=1-\left(\dfrac{1}{2}\right)^n[/tex]
Zatem
[tex]\displaystyle\\\Large\begin{aligned}\lim_{n\to\infty}\sqrt 2\cdot \sqrt[4]2\cdot \sqrt[8]2\cdot\ldots\cdot\sqrt[2^n]2=\lim_{n\to\infty}2^{1-\left(\frac{1}{2}\right)^n}=2^{1-0}=2^1=2\end{aligned}[/tex]