<s>[INST] Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
 [/INST] 
O method of differences, so powerful and yet so despised...

The $n$th term of the series to be computed is
$$
\prod_{k=1}^n\frac{2k+1}{2k+4}=\frac4{2n+4}\prod_{k=1}^n\frac{2k+1}{2k+2}=4(1-a_{n+1})\prod_{k=1}^na_k
$$ where $$a_k=\frac{2k+1}{2k+2}$$
By telescoping, each partial sum of the series is
$$
\sum_{n=0}^{N-1}\prod_{k=1}^n\frac{2k+1}{2k+4}=4-4\prod_{k=1}^Na_k
$$
Since $1-a_k\sim1/(2k)$, the product $\prod\limits_na_n$ diverges to $0$ hence the sum of the full series is $4$.

$$
\sum_{n=0}^\infty\prod_{k=1}^n\frac{2k+1}{2k+4}=4
$$

More generally, for every $(a,b)$ such that $a>-1$ and $b>a+1$, 
$$\sum_{n=0}^{N-1}\prod_{k=1}^n\frac{k+a}{k+b}=\frac{b}{b-a-1}\left(1-\prod_{k=1}^Na_k\right)$$ where $$a_k=\frac{k+a}{k+b-1}$$
hence, if furthermore $b\leqslant 2+a$, then the product $\prod\limits_na_n$ diverges to $0$ hence

$$\sum_{n=0}^\infty\prod_{k=1}^n\frac{k+a}{k+b}=\frac{b}{b-a-1}$$

The question above asks about the case $$a=1/2\qquad b=2$$ which fits these conditions.
Edit: Another exact formula for the partial sums, equivalent to the one above, is
$$
\sum_{n=0}^{N-1}\prod_{k=1}^n\frac{2k+1}{2k+4}=4-4\cdot\frac1{4^N}{2N+1\choose N}
$$
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