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Is your summation,\[\sum_{k=1}^{\infty}\left( \frac{1}{2^k}-\frac{1}{2^{k+1}} \right)\]?

So for the first N summations, you would have,\[\sum_{k=1}^{N}:=\frac{1}{2}-\frac{1}{2^{N+1}}\]

The series sums to 1/2.

thanks! so i this would be geometric, a = 1/2 and r= 1/2 ?

No, this is telescoping.

wonderful. thanks for your help! i'm still not looking forward to this exam tomorrow...

Good luck :)