I am recently struck upon this question that asks to find the sum until infinite terms $$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+.....∞$$ I tried my best to get something telescoping or something useful, but I failed. I even made a recurrence as $t_n=t_{n-1}\frac{2n-1}{2n+2}$, but this question was expected to be done with simple logic of series (Also that the recurrence on solving gives a higher order charasteristic polynomial, which might be difficult to solve without calculator). So, thus I ended up being confused with this question. So, can anyone prode a small solution to this (might be easy) problem .


While investigating the question, I found an easy answer., via telescoping series.: rearrange the fractions as $$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots=\frac{1}{2}=\frac{1\cdot(4-3)}{2\cdot4}+\frac{1\cdot3\cdot(6-5)}{2\cdot4\cdot6}+\cdots$$ $$\frac{1\cdot4}{2\cdot4}-\frac{1\cdot3}{2\cdot4}+\frac{1\cdot3\cdot6}{2\cdot4\cdot6}-\frac{1\cdot3\cdot5}{2\cdot4\cdot6}\cdots=\frac{1}{2}$$

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    $\begingroup$ Excellently done. +1. $\endgroup$ – Balarka Sen Sep 18 '14 at 13:47
  • $\begingroup$ Clean and concise. My answer pales in comparison. (+1) $\endgroup$ – SuperAbound Sep 18 '14 at 13:48

$$\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots=\frac{1}{2}$$

Rewrite the sum as \begin{align} \frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\cdots &=\sum^\infty_{n=0}\frac{(2n+1)!!}{(2n+4)!!}\\ &=\sum^\infty_{n=0}\frac{(2n+1)!}{2^{2n+2}n!(n+2)!}\\ &=\sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2}\Bigg{|}_{x=1/2} \end{align} Differentiate once to get \begin{align} \frac{{\rm d}}{{\rm d}x}\sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2} &=\sum^\infty_{n=0}\binom{2n+2}{n}x^{2n+1}\\ &=\frac{1}{i2\pi}\oint_{|z|=1}\frac{x(1+z)^2}{z}\sum^\infty_{n=0}\left[\frac{(x+xz)^{2}}{z}\right]^n{\rm d}z\\ &=\frac{1}{i2\pi}\oint_{|z|=1}\frac{x(1+z)^2}{z-x^2(1+z)^2}{\rm d}z\\ &=\frac{1}{i2\pi}\oint_{|z|=1}\frac{x(1+z)^2}{-x^2z^2+(1-2x^2)z-x^2}{\rm d}z\\ &=-\frac{1}{i2\pi x^2}\oint_{|z|=1}\frac{x(1+z)^2}{(z-r_+)(z-r_-)}{\rm d}z\\ &=-\frac{1}{x^2}{\rm Res}\left[\frac{x(1+z)^2}{(z-r_+)(z-r_-)},r_+\right]\\ &=-\frac{1}{x}\frac{(1+r_+)^2}{r_+ - r_-} \end{align} We know that $r_+=\dfrac{1-2x^2-\sqrt{1-4x^2}}{2x^2}$. Coupled with the fact that $r_+ - r_- =\dfrac{-\sqrt{1-4x^2}}{x^2}$, this gives us \begin{align} \sum^\infty_{n=0}\binom{2n+2}{n}x^{2n+1} &=\frac{\sqrt{1-4x^2}}{4x^3}-\frac{1}{2x^3}+\frac{1}{4x^3\sqrt{1-4x^2}} \end{align} Now, integrate this expression once \begin{align} \sum^\infty_{n=0}\frac{1}{2n+2}\binom{2n+2}{n}x^{2n+2} &=\frac{1-\sqrt{1-4x^2}}{4x^2}+C \end{align} It is easy to see that \begin{align} C =\lim_{x\to 0}\frac{\sqrt{1-4x^2}-1}{4x^2} =\lim_{u\to 1}\frac{u-1}{1-u^2} =-\lim_{u\to 1}\frac{1}{1+u} =-\frac{1}{2} \end{align} Finally, letting $x=\dfrac{1}{2}$ yields \begin{align} \sum^\infty_{n=0}\frac{(2n+1)!!}{(2n+4)!!} &=\frac{1-\sqrt{1-1}}{1}-\frac{1}{2}\\ &=\frac{1}{2} \end{align}

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    $\begingroup$ Actually, the result follows almost immediately if you recognize the series at line $3$ as Lagrange-Burmann inversion for $f(z)=(1+z)^2$ or some such. $\endgroup$ – Balarka Sen Sep 18 '14 at 13:35
  • $\begingroup$ @BalarkaSen That is interesting to know. I was unaware of the existence of such a formula and it seems like I used a sledgehammer to kill a fly. $\endgroup$ – SuperAbound Sep 18 '14 at 13:48

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