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

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?

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

## 2 Answers

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

Use

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

Rewrite the sum as

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

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