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oh you lazy
You could probably just sequentially find common denominators between the first terms of the sequence, compute the sum or difference of those numbers, and then move on to the next number carrying the number you found, repeat. e.g. 1 - 5/6 6/6 - 5/6 1/6 (1/6) + 7/12 2/12 + 7/12 9/12 and so on

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3/5
Dear santistebanc, I already know the answer, I just don't know how to do it in a faster way, that's all.
|dw:1333869325834:dw| maybe i'll just use my calculator... sorry...
Set up the summation: \[1+\sum_{n=2}^{9} \frac {(1+2n)(-1)^{n+1}} {(n+1)n} \]
From there you can use the Alternating Series Estimation Theorem.

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