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jmc

algebra intermediate

Problem

Compute

n = 1 \sum 1000 \frac{1}{n ^{2} + n} .

Solution

We can write

\frac{1}{n ^{2} + n} = \frac{( n + 1 ) - n}{n ( n + 1 )} = \frac{1}{n} - \frac{1}{n + 1} .

Thus, the sum telescopes:

n = 1 \sum 1000 \frac{1}{n ^{2} + n} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{2}{3}) + \dots + (\frac{1}{1000} - \frac{1}{1001}) = \frac{1}{1} - \frac{1}{1001} = \frac{1000}{1001} .

Final answer

\frac{1000}{1001}