# Examining convergence of the series. [closed]

Examine convergence of the series $$1-\frac{1}{2}+\frac{1\cdot3}{2\cdot4}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6}+\cdots$$

• Use Stirling's approximation. Jul 13, 2020 at 5:08
• @KaviRamaMurthy: I think things are much easier here ;) Jul 13, 2020 at 5:13
• What have you tried so far? Where are you stuck? Jul 13, 2020 at 5:20

Let's denote $$a_n=\frac{1 \cdot 3 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot \cdots \cdot (2n)}$$ and consider: $$\frac{a_n}{a_{n+1}}=1+\frac{1}{2n+1}$$ So by Gauss test we have absolute divergence and conditional convergence by Leibniz test and for limit 0
• For the Leibniz test we need to show $a_n\to0$. I don't see how this follows from the above. Jul 13, 2020 at 5:34
$$a_n=(-1)^n\frac{\prod_{k=1}^n (2k-1) } {\prod_{k=1}^n (2k) }=\binom{-\frac{1}{2}}{n}$$ and then $$S=\sum_{n=0}^\infty a_n x^n=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n} x^n=\frac{1}{\sqrt{1+x}}$$