Margfeldið
$$\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)
\left(1-\frac{1}{4^2}\right)\cdots
\left(1-\frac{1}{199^2}\right)\left(1-\frac{1}{200^2}\right)$$
er jafnt og
Lausn
Athugum að
$$1-\frac{1}{n^2}=\left(1-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)
=\frac{n-1}{n}\cdot\frac{n+1}{n}.$$
Því er margfeldið jafnt margfeldinu
$$
\begin{aligned} &\frac{2-1}{2}\cdot\frac{2+1}{2}\cdot \frac{3-1}{3}\cdot
\frac{3+1}{3}\cdots\frac{199-1}{199}\cdot\frac{199+1}{199}\cdot
\frac{200-1}{200}\cdot \frac{200+1}{200}\\ =\;&
\frac12\cdot\frac32\cdot\frac23\cdot\frac43\cdots\frac{198}{199}\cdot\frac{200}{199}\cdot\frac{199}{200}\cdot\frac{201}{200}
=\frac{1}{2}\cdot
\frac{201}{200}=\frac{201}{400}.\end{aligned}
$$