Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $

Question

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.

Answer 1

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 $\displaystyle(1+z)^\alpha$

Answer 2

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

Answer 3

Here is another approach. The answer equals $f(1)$ where $$\begin{align}f(x)&=1+\frac13x+\frac{1\cdot 3}{3\cdot 6}x^2+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^3+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}x^4+\cdots\\ \frac{df}{dx}&=\frac13\cdot 1+\frac{1\cdot 3}{3\cdot 6}x\cdot 2+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^2\cdot 3+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}x^3\cdot 4+\cdots\\ 3\frac{df}{dx}&=1+\frac13x\cdot 3+\frac{1\cdot 3}{3\cdot 6}x^2\cdot 5+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^3\cdot 7+\cdots\\ &=f(x)+2x\left(\frac13\cdot 1+\frac{1\cdot 3}{3\cdot 6}x\cdot 2+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}x^2\cdot 3+\cdots\right)\\ 3\frac{df}{dx}&=f(x)+2x\frac{df}{dx}\end{align}$$ Now solve the initial value problem with $f(0)=1$. $$\begin{align}(3-2x)\frac{df}{dx}&=f \\ \frac{df}{f}&=\frac{dx}{3-2x} \\ \ln f&=\ln [C(3-2x)^{-1/2}]\\ f(x)&=C(3-2x)^{-1/2}\\ f(0)&=1=C\cdot 3^{-1/2} \\ C&=3^{1/2} \\ f(x)&=3^{1/2}(3-2x)^{-1/2}\\ f(1)&=3^{1/2}.\end{align}$$

Answer 4

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 $T=\sqrt{3}$.

EDIT

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.

Answer 5

\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*}

Answer 6

HINT:

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

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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) $\sqrt{3}$

Answer 7

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.