calculus - How to evaluate the infinite series: $frac 1 {3cdot6} + frac 1 {3cdot6cdot9} +frac 1 {3cdot6cdot9cdot12}+ldots$

The infinite series is given by:

$$\dfrac 1 {3\cdot6} + \dfrac 1 {3\cdot6\cdot9} +\dfrac 1 {3\cdot6\cdot9\cdot12}+\ldots$$

What I thought of doing was to split the general term as:

$$\begin{align} t_r &= \dfrac 1 {3^{r+1}(r+1)!}\\\\ &= \dfrac {r+1 - r} {3^{r+1}(r+1)!}\\\\ &= \dfrac {1} {3^{r+1}\cdot r!} - \dfrac{r}{3^{r+1}(r+1)!} \end{align}$$

But this doesn't seem to help.

HINTS?

Answer

HINT:

$$e^x=\sum_{0\le r<\infty}\frac{x^r}{r!}$$

Can you take it from here?

A strongly resembling sequence $$-\ln(1-x)=\sum_{1\le r<\infty}\frac{x^r}{r}$$ for $-1\le x<1$