<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] A probability approach. 
Let $\{C_n\}_{n=1}^{\infty}$ be independent random bits such that $P(C_n=1)=\frac{1}{2(n+1)}$. 
Then $$\prod_{k=1}^{n}P(C_k=0)=\prod_{k=1}^{n}\left(1-\frac{1}{2(k+1)}\right)\to 0\text{ as }n\to\infty$$
So we can define a random variable $X$ to be the least $n$ such that $C_n=1$, and you get that $$\sum_{k=1}^{\infty} P(X=n) = 1,$$ since $P(X>n)\to 0$.
But a quick calculation shows:
$$P(X=n)=P(C_n=1)\prod_{k=1}^{n-1}P(C_k=0)=\frac{1}{2(n+2)}\prod_{k=1}^{n-1} \frac{2k+1}{2k+2} = \frac{1\cdot 3\cdot \cdots \cdot (2n-1)}{4\cdot 6\cdot \cdots \cdot 2(n+2)}$$
Now your sum is $4\sum_{n=1}^{\infty} P(X=n) = 4$.
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