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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?

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    $\begingroup$ $$\prod_{j=1}^i \frac{1}{2j}=\frac{2^{-i}}{i!}$$ $\endgroup$
    – saulspatz
    Jan 2, 2020 at 19:58
  • $\begingroup$ that makes sense $\endgroup$
    – user738928
    Jan 2, 2020 at 19:59

2 Answers 2

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

Use

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

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Rewrite the sum as

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

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

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