Sum of series : $$1+\frac{1}{4}+\frac{1\times3}{4\times8}+\frac{1\times3\times5}{4\times8 \times12}+\cdots$$

$n$-th term of series is $$a_{n} =\frac{1\times3 \times5\times\cdots \times(2n-1)}{4\times8\times12\times \cdots \times4n} = \prod^{n}_{k=1}{2k-1\over4k}$$

I can not go further,

  • $\begingroup$ Is the answer $\sqrt{2} $ $\endgroup$ – Archis Welankar Oct 8 '16 at 14:56
  • $\begingroup$ @ArchisWelankar: Alpha says so I would look to see the series as a Taylor series for a trig function, but haven't found it. $\endgroup$ – Ross Millikan Oct 8 '16 at 15:06
  • $\begingroup$ @Ross Millikan can you please check it again as my answer matches with answer below $\endgroup$ – Archis Welankar Oct 8 '16 at 16:14
  • $\begingroup$ @ArchisWelankar: Why check again? I agreed with you, just did not know how to derive it. I got to line $(2)$ of the answer below. $\endgroup$ – Ross Millikan Oct 8 '16 at 19:20
  • $\begingroup$ @MarkusScheuer Got the point and the error. I preserved everything original, so OP does not need to know product symbol. Also posting this type of problem and not knowing product symbol is rare. I think he does not know the Latex command for the symbol. $\endgroup$ – A---B Oct 8 '16 at 20:56

Note, the general term $a_n, n\geq 1$ is \begin{align*} a_n=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{4\cdot8\cdot 12\cdots (4n)} =\frac{1\cdot 3\cdot 5\cdots (2n-1)}{(1\cdot 2\cdot 3\cdots n)\cdot 4^n} =\frac{(2n-1)!!}{n!}\cdot \frac{1}{4^n}\\ \end{align*}

with $$(2n-1)!!=(2n-1)\cdot(2n-3)\cdots 5\cdot 3\cdot 1$$ double factorials.

We obtain \begin{align*} 1+\sum_{n=1}^\infty\frac{(2n-1)!!}{n!}\cdot\frac{1}{4^n} &=\sum_{n=0}^\infty\frac{(2n)!}{n!(2n)!!}\cdot\frac{1}{4^n}\tag{1}\\ &=\sum_{n=0}^\infty\frac{(2n)!}{n!n!2^n}\cdot\frac{1}{4^n}\tag{2}\\ &=\sum_{n=0}^\infty\binom{2n}{n}\frac{1}{8^n}\tag{3}\\ &=\left.\frac{1}{\sqrt{1-4z}}\right|_{z=\frac{1}{8}}\tag{4}\\ &=\frac{1}{\sqrt{1-\frac{1}{2}}}\\ &=\sqrt{2} \end{align*}


  • In (1) we apply $(2n)!=(2n)!!\cdot (2n-1)!!$.

  • In (2) we use $(2n)!!=n!2^n$.

  • In (3) we use $\binom{2n}{n}=\frac{(2n)!}{n!n!}$.

  • In (4) we use the generating function of the central binomial coefficient.

Add-on [2017-03-12] according to a comment from @navinstudent.

Using the binomial series expansion we obtain \begin{align*} \frac{1}{\sqrt{1-4z}}&=(1-4z)^{-\frac{1}{2}}=\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-4z)^n\qquad\qquad |z|<\frac{1}{4} \end{align*} Since \begin{align*} \binom{-\frac{1}{2}}{n}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)\cdots\left(-\frac{1}{2}-n+1\right)}{n!}\\ &=\frac{1}{n!}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\cdots\left(-\frac{2n-1}{2}\right)\\ &=\frac{(-1)^n}{2^nn!}\cdot(2n-1)!! =\frac{(-1)^n}{2^nn!}\cdot\frac{(2n)!}{(2n)!!} =\frac{(-1)^n}{2^nn!}\cdot\frac{(2n)!}{2^nn!}\\ &=\frac{(-1)^n}{4^n}\binom{2n}{n} \end{align*} we get $$\sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-4z)^n=\sum_{n=0}^\infty \binom{2n}{n}z^n$$ and step (4) follows.

  • $\begingroup$ If I may add, for $k>2$, then $$\sum_{n=0}^\infty\binom{2n}{n}\frac{1}{(2^k)^n}=\sqrt{\frac{2^{k-2}}{2^{k-2}-1}}$$ where the OP's was just the case $k=3$. $\endgroup$ – Tito Piezas III Oct 15 '16 at 4:36
  • $\begingroup$ @TitoPiezasIII: Interesting! Thanks for your comment! $\endgroup$ – Markus Scheuer Oct 18 '16 at 12:51
  • $\begingroup$ Can you please provide a prove of generating function you used in point 4 $\endgroup$ – Navin Mar 12 '17 at 8:14
  • $\begingroup$ @navinstudent: Explanation added. $\endgroup$ – Markus Scheuer Mar 12 '17 at 8:57
  • $\begingroup$ @navinstudent: You're welcome! :-) $\endgroup$ – Markus Scheuer Mar 12 '17 at 11:07

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