Help with telescoping sum $\sum_{i=3}^n \frac{1}{i(i+3)} $

real-analysis sequences-and-series summation telescopic-series

$\sum_{i=3}^{n} \frac{1}{i(i+3)} = \frac{1}{3} \sum_{i=3}^{n} \frac{1}{i} - \frac{1}{i+3}$ via partial fractions.

To see what's going on, try writing out the first 4 or so terms and you'll quickly see what cancels out and what's left over...

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