91- 6303239042
For $$\sum_{k=1}^{n} \frac{1}{k(k+1)}$$:
$$ \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} $$
Sum: $$ 1 - \frac{1}{n+1} = \frac{n}{n+1} $$
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