Why is $\sum\limits_{i=1}^\infty (-1)^{i-1}\prod\limits_{j=1}^i \frac{1}{2j}=1-\frac1{\sqrt{e}}$?

Question

Why is $\displaystyle\sum_{i=1}^\infty (-1)^{i-1}\displaystyle\prod_{j=1}^i \frac{1}{2j}=1-\frac1{\sqrt{e}}$?

It makes sense that the series converges (it's an alternating series where $a_n$ is positive and decreasing), but I don't know why it would converge to that value. Perhaps recalling the Taylor series of $e^x$ would be useful?

Answer 1

The left side $$=\sum_{r=1}^\infty\dfrac{(-1/2)^r}{r!}$$

Use

$$\sum_{r=0}^\infty\dfrac{x^r}{r!}=e^x$$

Answer 2

Rewrite the sum as

$$\sum_{n\geqslant 1}{\left(\frac{-1}{2}\right)}^n \frac1{n!}$$

and consider the power series for $\exp(x)$.