How the telescopic sum $$1+\sum _{n=1}^{\infty } \dfrac{(2 n+1)!}{2^{2 n-1} n! (n+2)!}$$ has value $4$?
I think this has something to do with Taylor series but I'm not able to get it. Actually, while looking up for Finding the $nth$ term of a sequence answer by Raffaele last step, there I've my confusion.
Method I - as a telescoping sum
First, we rewrite the sum at hand in terms of gamma functions $$\mathcal{S} \stackrel{def}{=}1 + \sum_{n=1}^\infty \frac{(2n+1)!}{2^{2n-1}n!(n+2)!} = 1 + \sum_{n=1}^\infty \frac{\Gamma(2(n+1))}{2^{2n-1}\Gamma(n+1)\Gamma(n+3)} $$ Using the duplication formula of gamma function, we find
$$\Gamma(z)\Gamma\left(z+\frac12\right) = 2^{1-2z}\sqrt{\pi}\Gamma(2z) \quad\implies\quad \Gamma(2(n+1)) = \frac{2^{2n+1}}{\sqrt{\pi}}\Gamma(n+1)\Gamma\left(n+\frac32\right)$$ This leads to $$\mathcal{S} = 1 + \frac{4}{\sqrt{\pi}}\sum_{n=1}^\infty \frac{\Gamma\left(n+\frac32\right)}{\Gamma(n+3)}$$
Notice for any $\alpha, \beta \in \mathbb{R}$ such that $\beta \ne \alpha+1$, we have
$$\begin{align}\frac{\Gamma(n+\alpha)}{\Gamma(n+\beta)} &= \frac{(n+\beta - 1) - (n+\alpha)}{\beta - (\alpha+1)}\frac{\Gamma(n+\alpha)}{\Gamma(n+\beta)}\\ &= \frac{1}{\beta - (\alpha+1)}\left[\frac{\Gamma(n+\alpha)}{\Gamma(n+\beta-1)} - \frac{\Gamma(n+\alpha+1)}{\Gamma(n+\beta)}\right] \end{align}\tag{*1}$$ The RHS has the right form for a telescoping sum. This means for any $m > 0$, we have
$$\sum_{n=1}^m \frac{\Gamma(n+\alpha)}{\Gamma(n+\beta)} = \frac{1}{\beta - (\alpha+1)}\left[\frac{\Gamma(1+\alpha)}{\Gamma(\beta)} - \frac{\Gamma(m+\alpha+1)}{\Gamma(m+\beta)}\right]$$
For any $\gamma \in \mathbb{C}$, it is known that $\displaystyle\;\lim_{n\to\infty} \frac{\Gamma(n+\gamma)}{\Gamma(n)n^\gamma} = 1$.
As a corollary, when $\beta > \alpha + 1$, $\displaystyle\;\lim_{m\to\infty}\frac{\Gamma(m+\alpha+1)}{\Gamma(m+\beta)} = 0$. For such $\alpha, \beta$, we have
$$\sum_{n=1}^\infty \frac{\Gamma(n+\alpha)}{\Gamma(n+\beta)} = \frac{1}{\beta - (\alpha+1)}\frac{\Gamma(1+\alpha)}{\Gamma(\beta)}$$
In particular, if we take $(\alpha,\beta) = (\frac32,3)$, we get $$\mathcal{S} = 1 + \frac{4}{\sqrt{\pi}}\times\frac{1}{3-(\frac32+1)}\times\frac{\Gamma(\frac52)}{\Gamma(3)} = 1 + \frac{4}{\sqrt{\pi}}\times 2 \times \frac{\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)\sqrt{\pi}}{2!} = 1 + 3 = 4$$
Method II - as an integral.
IMHO, instead of casting $\mathcal{S}$ to a telescoping sum, an easier way (for most human) is convert $\mathcal{S}$ to an integral. Recall following formula for the OGF of centeral binomial coefficients.
$$\sum_{n=0}^\infty \binom{2n}{n}t^n = \frac{1}{\sqrt{1-4t}}\tag{*2}$$
We can evaluate the sum at hand as
$$\begin{align}\mathcal{S} &= 1 + \sum_{n=1}^\infty \frac{(2n+2)!}{4^n(n+1)!^2(n+2)} = 1 + 4\sum_{n=2}^\infty \binom{2n}{n} \frac{1}{4^n(n+1)}\\ &= 1 + 4\sum_{n=2}^\infty \binom{2n}{n}\int_0^1\left(\frac{t}{4}\right)^n dt = 1 + 4\int_0^1 \sum_{n=2}^\infty \binom{2n}{n}\left(\frac{t}{4}\right)^n dt\\ &= 1 + 4\int_0^1 \left( \color{red}{\sum_{n=0}^\infty \binom{2n}{n}\left(\frac{t}{4}\right)^n} - 1 - \frac{t}{2}\right)dt\\ &= 1 + 4\int_0^1\left(\color{red}{\frac{1}{\sqrt{1-t}}} - 1 - \frac{t}{2}\right)dt = 4\end{align}$$
TL;DR
Finally, I will like to make some comments on Method I. As one can see from above derivation, if one knows the relation $(*1)$, turning the sum to a telescoping one is relatively straight forward. The real problem is for human, it is not that easy to detect the pattern embedded in $(*1)$.
For many sums involving binomial coefficients, factorials and in general any hypergeometric series, there is something called Wilf-Zeilberger pair which may be used to find an equivalent and much simpler sum. In most cases, finding WZ pairs by hand is impractical. However, there are algorithms which allow machine to find one (if it exists) for us!
Some CAS has implemented those algorithms and that's why they can simplify those nasty looking finite sums into a much simpler one. If you want to turn similar looking sums into a telescoping one, I'll recommend you first look at your CAS and see whether it has this sort of algorithms implemented!
$$1 + \sum {n=1}^\infty \frac{(2 n+1)!}{2^{2 n-1} n! (n+2)!} = 1 + \sum{n=1}^\infty \frac{(2n+2)!}{4^n(n+1)!^2(n+2)} = 1 + 4\sum_{n=2}^\infty \binom{2n}{n} \frac{1}{4^n(n+1)}\ = 1 + 4\int_0^1 \sum_{n=2}^\infty \binom{2n}{n}\left(\frac{t}{4}\right)^n dt = 1 + 4\int_0^1\left(\frac{1}{\sqrt{1-t}} - 1 - \frac{t}{2}\right)dt = 4 $$
– achille hui Dec 23 '17 at 14:32