(1/(1+2))+(1/(1+2+3))+(1/(1+2+3+4))+...+(1/(1+2+3+4+...+29))


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Question Number 105526 by bramlex last updated on 29/Jul/20

$\frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + \frac{1}{1 + 2 + 3 + 4} + . . . + \frac{1}{1 + 2 + 3 + 4 + . . . + 29}$
	Answered by Dwaipayan Shikari last updated on 29/Jul/20

	$(1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + . . . . + \frac{1}{1 + 2 + 3 + 4 + 5 + . . + 29}) - 1$ $T_{n} = \frac{1}{\frac{n (n + 1)}{2}} = \frac{2}{n (n + 1)}$ $\sum^{n} T_{n} = 2 Σ \frac{1}{n} - \frac{1}{n + 1}$ $\sum^{29} T_{29} = 2 (1 - \frac{1}{30}) = \frac{29}{15}$ $S o a n s w e r i s (\sum^{29} T_{29} - 1) = \frac{14}{15}$
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