For example, how can I find the sum of $\frac 1 {(2 + n) \cdot (3 + n)}$ + ... for $n = 0$ to $n = 97$ i.e.
$$\frac{1}{2 \cdot 3} + \frac 1 {3 \cdot 4} + ... + \frac 1 {99 + 100}$$
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For some sums like this there is a trick that allows you to convert the sum to a known sum. There is a trick here – Alexander Vlasev Sep 28 '13 at 04:37
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@SujaanKunalan My bad. Deleting the comment. – Memming Sep 28 '13 at 04:42
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Thank you for getting the parentheses right. This is rarer than it should be. – Ross Millikan Sep 28 '13 at 04:43
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Note that $$\frac{1}{n(n + 1)} = \frac{1}{n} - \frac{1}{n + 1}$$
Hence, your second series can be rewritten as
$$\frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ...$$
terminating, of course, at an appropriate point.
Similarly,
$$\frac{1}{(2 + n)(3 + n)} = \frac{1}{2 + n} - \frac{1}{3 + n}$$
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@Jason: It's worth noting in general that there is no way to find such short formulae for summations in terms of just the boundary values (in your case 0 and 97). In many others there might be a method, but it is extremely difficult. – Eric Stucky Dec 24 '13 at 08:37