Discussion

Explanation:

N = $\frac{1}{1\times 2}$ + $\frac{1}{2\times 3}$ + $\frac{1}{3\times 4}$ + ... + $\frac{1}{n\times (n+1)}$

N = $\frac{2-1}{1\times 2}$ + $\frac{3-2}{2\times 3}$ + $\frac{4-3}{3\times 4}$ + ... + $\frac{(n+1)-n}{n\times (n+1)}$

N = $\frac{2}{1\times 2}$ - $\frac{1}{1\times 2}$ + $\frac{3}{2\times 3}$ - $\frac{2}{2\times 3}$ + $\frac{4}{3\times 4}$ - $\frac{3}{3\times 4}$ + ... + $\frac{(n+1)}{n\times (n+1)}$ - $\frac{n}{n\times (n+1)}$

N = $\frac{1}{1}$ - $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ + $\frac{1}{3}$ - $\frac{1}{4}$ + ... + $\frac{1}{n}$ - $\frac{1}{(n+1)}$

N = $\frac{1}{1}$ - $\frac{1}{n}$

Since 'n' is a natural numbers N will always be less than 1.

Hence, option (a).

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