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 $-1\le x<1$
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