How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $$

I can see that 3 cancels out after 1/3, but what next? I can't go further.

  • $\begingroup$ [WolframAlpha](wolframalpha.com/input/… says $\sqrt{3}$. $\endgroup$ – Arthur Apr 9 '14 at 8:16
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    $\begingroup$ This is a binomial series. $\endgroup$ – Lucian Apr 9 '14 at 8:39
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    $\begingroup$ The general term seems to be $\frac {\prod_1^n 2k-1} {\prod_1^n 3k}=\frac{(2n)!}{3^n2^n(n!)^2}=\binom{2n}{n}(\frac 1 6)^n$ $\endgroup$ – pmichel31415 Apr 12 '16 at 18:17

As the denominator of the $n$th term $T_n$ is $\displaystyle3\cdot6\cdot9\cdot12\cdots(3n)=3^n \cdot n!$

(Setting the first term to be $T_0=1$)

and the numerator of $n$th term is $\displaystyle1\cdot3\cdot5\cdots(2n-1)$ which is a product of $n$th terms of an Arithmetic Series with common difference $=2,$

we can write $\displaystyle1\cdot3\cdot5\cdots(2n-1)=-\frac12\cdot\left(-\frac12-1\right)\cdots\left(-\frac12-{n+1}\right)\cdot(-2^n)$

which suitably resembles the numerator of Generalized binomial coefficients

$$\implies T_n=\frac{-\frac12\cdot\left(-\frac12-1\right) \cdots\left(-\frac12-{n+1}\right)}{n!}\left(-\frac23\right)^n$$

So, here $\displaystyle z=-\frac23,\alpha=-\frac12$ in $\displaystyle(1+z)^\alpha$


Using Generalized Binomial Expansion, $$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$ given the converge holds

Comparing with given Series $\displaystyle nx=\frac13\implies n^2x^2=\cdots\ \ \ \ (1)$

and $\displaystyle\frac{n(n-1)}{2!}x^2=\frac{1\cdot3}{3\cdot6}\ \ \ \ (2)$

Divide $(2)$ by $(1)$ to find $\displaystyle n=-\frac12$ and consequently $\displaystyle x=-\frac23$

Observe that these values satisfy the next two terms, too.

Hence, the sum follows


Consider denominator and numerator separately at first,

$$G_n = 2^n \prod_{m=1}^n m-1/2, \qquad F_n = \frac{1}{3^n n!}$$

Thus we have

$$T_n = \prod_{m=1}^n \frac{m-1/2}{n!} \left(\frac{2}{3}\right)^n \qquad \text{or} \qquad T = \sum_{n=0}^\infty \prod_{m=1}^n \frac{m-1/2}{n!}\left(\frac{2}{3}\right)^n$$ Looking these series elements up we arrive at $T=\sqrt{3}$.


The final form of the series is $$\sum_{n=0}^\infty \frac{\Gamma (n+1/2) }{\sqrt{\pi} n!} \left( \frac{2}{3}\right)^n =\sqrt{3}$$ where $\Gamma(n)$ is the well-known Gamma function.

  • $\begingroup$ Thank you.☺ I appreciate the notations. $\endgroup$ – Silent Apr 9 '14 at 9:31
  • $\begingroup$ I have just slightly changed the notation. $\endgroup$ – Caelesticum Apr 9 '14 at 9:33

\begin{align*} 1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+ \frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+\ldots &=\sum_{n=0}^{\infty} \frac{(2n-1)!!}{3^{n} n!} \\ &=\sum_{n=0}^{\infty} \frac{(-\frac{1}{2})(-\frac{3}{2})\ldots (-\frac{2n-1}{2})} {3^{n} n!} (-2)^{n} \\ &=\sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} \frac{ \left( -\frac{2}{3} \right)^{n}} {n!} \\ &= \left[ 1+\left( -\frac{2}{3} \right) \right]^{-\frac{1}{2}} \\ &= \sqrt{3} \end{align*}



Look at the Taylor series of $f(x)=(1-x)^{-1/2}$ evaluated at $x=1/3$.

Note that the general terms of the series can be written $$a_n=\frac{(2n-3)!!}{3^{n-1}(n-1)!}$$The series can be written as $$1+\sum_{n=1}^\infty \frac{(2n-1)!!}{3^n\,n!}=\sum_{n=0}^\infty \frac{(2n)!}{4^n\,(n!)^2}\left(\frac23\right)^n$$which we recognize as the series for $(1-x)^{-1/2}$ evaluated at $x=2/3$. Therefore, the answer is (c) $\sqrt{3}$

  • $\begingroup$ It will be great , if you could elucidate the thought process that led you to think of the function $ (1-4x)^{1/2} $ in the first place . $\endgroup$ – Noob101 Apr 12 '16 at 18:23
  • $\begingroup$ Actually, it is easier just to recognize the series form for $(1-x)^{-1/2}$. $\endgroup$ – Mark Viola Apr 12 '16 at 18:39

From the OGF of Catalan numbers we have that: $$ \sum_{n\geq 0}\binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}} $$ where the radius of convergence of the LHS is $\frac{1}{4}$ since $\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}=\frac{4n+2}{n+1}$.

By evaluating the previous identity at $x=\frac{1}{6}$ it follows that: $$ \color{red}{\sqrt{3}} = 1+\sum_{n\geq 1}\frac{(2n)!}{n! n! 6^n} = 1+\sum_{n\geq 1}\frac{(2n-1)!!}{3^n n!} $$ where the RHS is exactly our sum.


Here is another approach. The answer equals $f(1)$ where $$f(x)=1+\frac13x+\frac{1.3}{3.6}x^2+\cdots\\ 3\frac{df}{dx}=1+\frac13x.3+\frac{1.3}{3.6}x^2.5+\cdots\\ =f(x)+2\left(\frac13x.1+\frac{1.3}{3.6}x^2.2+\cdots\right)\\ 3\frac{df}{dx}=f(x)+2x\frac{df}{dx}$$ Now solve the differential equation.


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