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The series $\sum_{i=1}^\infty2^{-i}/i!$ is clearly convergent by the ratio test with $\sum_{i=1}^{\infty}2^{-i}$, but is it possible to calculate the exact sum?

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    $\begingroup$ $e^{1/2}-1$ is it. $\endgroup$ – Daniel Fischer Oct 22 '13 at 14:07
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More generally, $$e^x=\sum_{i=0}^\infty \frac{1}{i!}x^i$$ Applying this with $x=2^{-1}$ gives the series you want, but starting at $i=0$. That is an extra $2^0/0!=1$ in the sum. Subtract this off and you get $\sqrt{e}-1$, as @Daniel points out in the comments.

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