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$$1 + \frac{1}{1+2} + \frac{1}{1+2+3} + \dots + \frac{1}{1+2+3+\dots+2006} =$$ ____________
Correct Answer: 1.999
The general term is $$\frac{1}{1+2+\dots+n} = \frac{2}{n(n+1)} = 2\left(\frac{1}{n} - \frac{1}{n+1}\right)$$, and the first term 1 is the case $$n = 1$$. Summing from $$n = 1$$ to $$n = 2006$$, the series telescopes to $$2\left(1 - \frac{1}{2007}\right) = \frac{4012}{2007} \approx 1.999$$.
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