TITLE: 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 [33 upvotes]: 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.

REPLY [10 votes]: 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 $f(0)=1$.
$$\begin{align}(3-2x)\frac{df}{dx}&=f \\
\frac{df}{f}&=\frac{dx}{3-2x} \\
\ln f&=\ln [C(3-2x)^{-1/2}]\\
f(x)&=C(3-2x)^{-1/2}\\
f(0)&=1=C\cdot 3^{-1/2} \\
C&=3^{1/2} \\
f(x)&=3^{1/2}(3-2x)^{-1/2}\\
f(1)&=3^{1/2}.\end{align}$$