Find the sum. Find the sum

$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{2002 \cdot 2003}$
It is easy to check the equality $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$.

Applying this equality for k from 1 to 2002, we write

$\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{2002 \cdot 2003}=$

$=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2002}-\frac{1}{2003}\right)=$

$=1-\frac{1}{2003}=\frac{2002}{2003}$.

Remark. The number 2003 can be replaced by any natural number.

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