# Does series$\sum_{n=0}^\infty (-1)^n {(2n)!\over 4^nn!(n+1)!}$ converges absolutely or conditionally?

I wonder the convergence of the series $$\sum_{n=1}^\infty (-1)^n {(2n)!\over 4^nn!(n+1)!}.$$

We can derive that the general term is equal to $$(-1)^n \times {1\over2} \times {3 \over 4} \times \cdots \times {2n-1 \over 2n} \times {1 \over n+1}.$$

We can easily see that the series derived by taking absolute value is monotone decreasing and converges to $0$, so that the original series converges, by Alternating Series Test. Then, does it converges absolutely/conditionally? I feel like it converges absolutely, since ${1\over n}$ is somewhat the threshold of convergence/divergence and the term ${1\over2} \times {3 \over 4} \times \cdots \times {2n-1 \over 2n}$ is sufficiently small that it will make the series converge, but I have no idea how to deal with it.

• What have you tried? Have you attempted to apply any of the usual tests for convergence (e.g. the ratio test, the root test, etc.)? – Xander Henderson Nov 22 '17 at 16:30
• I tried both ratio test and root test, but did not work. maybe I should try to apply abel/dirichlet test?? they are the most powerful tests that the textbook serves – Jeongmin Park Nov 22 '17 at 16:32

Yes, it converges absolutely. By using Stirling approximation $n!=\sqrt{2\pi n}(n/e)^n$ (see also the approximation of the central binomial coefficient here), it follows that $${(2n)!\over 4^nn!(n+1)!}=\frac{1}{n+1}\cdot\frac{\binom{2n}{n}}{4^n}\sim\frac{1}{n}\cdot\frac{1}{\sqrt{\pi n}}\sim\frac{1}{\sqrt{\pi}n^{3/2}}$$ and the series $\sum 1/n^{3/2}$ is convergent because $3/2>1$.
You may notice that the extended binomial theorem gives (for any $x\in(-1,1)$): $$\frac{1}{\sqrt{1-x}}=(1-x)^{-1/2} = \sum_{n\geq 0}\binom{-\tfrac{1}{2}}{n}(-1)^n x^n =\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}x^n$$ hence by performing $\int_{0}^{z}(\ldots)\,dx$, we get: $$2-2\sqrt{1-z} = \sum_{n\geq 0}\frac{\binom{2n}{n}}{(n+1)4^n}\,z^{n+1}$$ then by dividing both sides by $z$ and replacing $z$ with $-x$: $$\frac{2}{1+\sqrt{1+x}} = \sum_{n\geq 0}\frac{\binom{2n}{n}(-1)^n}{(n+1)4^n}\,x^n$$ and subtracting $1$ to both sides: $$\frac{1-\sqrt{1+x}}{1+\sqrt{1+x}} = \sum_{n\geq 1}\frac{\binom{2n}{n}(-1)^n}{(n+1)4^n}\,x^n.$$ Since
$$\frac{1}{4^n}\binom{2n}{n}=\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta \leq \frac{2}{\pi}\int_{0}^{\pi/2}e^{-n\theta^2}\,d\theta\leq \frac{2}{\pi}\int_{0}^{+\infty}e^{-n\theta^2}\,d\theta = \frac{1}{\sqrt{\pi n}}$$ the previous series evaluated at $x=1$ is absolutely convergent and $$\sum_{n\geq 1}\frac{\binom{2n}{n}(-1)^n}{(n+1)4^n} = \frac{1-\sqrt{2}}{1+\sqrt{2}} = \color{red}{2\sqrt{2}-3}.$$
With Gauss' test: $$\Big|\dfrac{a_{n+1}}{a_n}\Big|=\dfrac{2n+1}{2n+4}=1-\dfrac{\frac32}{n}+\dfrac{\frac{3n}{n+2}}{n^2}$$ and $\dfrac{3n}{n+2}<3$, so with $p=\dfrac32>1$ the series is converges absolutely.