Find the limit of the sequence:
$$a_n= \left ( 1- \frac{1}{2^2} \right )\left ( 1- \frac{1}{3^2} \right )\cdots \left ( 1- \frac{1}{n^2} \right ), \;\; n \geq 2$$
Solution
The sequence takes its equivalent form:
$$\begin{aligned}
\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ) &=\prod_{k=2}^{n}\left [ \left ( 1- \frac{1}{k} \right )\left ( 1+ \frac{1}{k} \right ) \right ] \\
&= \left [ \left ( 1- \frac{1}{2} \right )\left ( 1- \frac{1}{3} \right ) \cdots \left ( 1- \frac{1}{n} \right ) \cdot \left ( 1+ \frac{1}{2} \right )\left ( 1+ \frac{1}{3} \right )\cdots \left ( 1+ \frac{1}{n} \right ) \right ] \\
&=\frac{1}{2}\cdot \frac{2}{3}\cdots \frac{n-1}{n}\cdot \frac{3}{2}\cdot \frac{4}{3}\cdots \frac{n+1}{n} \\
&=\frac{1}{2}\cdot \frac{n+1}{n} \xrightarrow{n \rightarrow +\infty}\frac{1}{2}
\end{aligned}$$
Hence the limit of the sequence is $1/2$.
$$a_n= \left ( 1- \frac{1}{2^2} \right )\left ( 1- \frac{1}{3^2} \right )\cdots \left ( 1- \frac{1}{n^2} \right ), \;\; n \geq 2$$
Solution
The sequence takes its equivalent form:
$$\begin{aligned}
\prod_{k=2}^{n}\left ( 1-\frac{1}{k^2} \right ) &=\prod_{k=2}^{n}\left [ \left ( 1- \frac{1}{k} \right )\left ( 1+ \frac{1}{k} \right ) \right ] \\
&= \left [ \left ( 1- \frac{1}{2} \right )\left ( 1- \frac{1}{3} \right ) \cdots \left ( 1- \frac{1}{n} \right ) \cdot \left ( 1+ \frac{1}{2} \right )\left ( 1+ \frac{1}{3} \right )\cdots \left ( 1+ \frac{1}{n} \right ) \right ] \\
&=\frac{1}{2}\cdot \frac{2}{3}\cdots \frac{n-1}{n}\cdot \frac{3}{2}\cdot \frac{4}{3}\cdots \frac{n+1}{n} \\
&=\frac{1}{2}\cdot \frac{n+1}{n} \xrightarrow{n \rightarrow +\infty}\frac{1}{2}
\end{aligned}$$
Hence the limit of the sequence is $1/2$.
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