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## Infinite product

### Solution

\begin{aligned} P_{n} &=\frac{e}{\sqrt{e}} \cdot \frac{\sqrt[3]{e}}{\sqrt[4]{e}} \cdot \frac{\sqrt[5]{e}}{\sqrt[6]{e}} \cdots \frac{\sqrt[n-1]{e}}{\sqrt[n]{e}} \\\\ &=\frac{e^{1}}{e^{\frac{1}{2}}} \cdot \frac{e^{\frac{1}{3}}}{e^{\frac{1}{4}}} \cdot \frac{e^{\frac{1}{5}}}{e^{\frac{1}{6}}} \cdots \frac{e^{\frac{1}{n-1}}}{e^{\frac{1}{n}}}\\\\ &=e^{1}\cdot e^{-\frac{1}{2}}\cdot e^{\frac{1}{3}}\cdot e^{-\frac{1}{4}} \cdot e^{\frac{1}{5}}\cdot e^{-\frac{1}{6}}\cdots e^{\frac{1}{n-1}}\cdot e^{-\frac{1}{n}}\\\\ &=e^{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{n-1}-\frac{1}{n}} \end{aligned}
$\text { Maclaurin Series of } \ln (1+x) :$
$\begin{gathered} \ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}+\cdots \\\\ -1<x\leq 1\\\\ For\; x=1\\\\ \ln (1+1)=1-\frac{1}{2}+\frac{1}{3}+\cdots\\\\ \ln (2)=1-\frac{1}{2}+\frac{1}{3}+\cdots \end{gathered}$
\begin{aligned} \lim_{n \rightarrow \infty}P_{n} &=\lim_{n \rightarrow \infty}e^{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{n-1}-\frac{1}{n}}\\\\ &=e^{\lim_{n \rightarrow \infty}1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{n-1}-\frac{1}{n}}\\\\ &=e^{\ln2}\\\\ &=2 \end{aligned}
$\large \Rightarrow\frac{e}{\sqrt{e}} \cdot \frac{\sqrt[3]{e}}{\sqrt[4]{e}} \cdot \frac{\sqrt[5]{e}}{\sqrt[6]{e}}\cdots=2$
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