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) $4$.

Note that
$$\binom{n}{r} = \frac{n(n-1)(n-2)\dots(n-r+1)}{1\cdot2\cdot3\cdots r}$$
and so
$$\binom{1/2}{r} = \frac{\frac12 \cdot \frac{-1}2 \cdot \frac{-3}{2} \cdot \frac{-5}{2} \cdots \frac{-(2r-3)}{2}}{1\cdot2\cdot3\cdots r} = \frac{(-1)^{r-1}1 \cdot 3 \cdot 5 \cdots (2r-3)}{2^r r!}$$
Now, in this problem, each term after the "$1$" term follows a pattern. Let the last factor in the denominator of a term be $2r$, then the general term is:
$$
\frac{1 \cdot 3 \cdot 5 \cdots (2r - 3)}{6 \cdot 8 \cdot 10 \cdots (2r)}
= \frac{1 \cdot 3 \cdot 5 \cdots (2r - 3)}{2^{r-2} 3 \cdot 4 \cdot 5 \cdots r}
= \frac{1 \cdot 3 \cdot 5 \cdots (2r - 3)}{2^{r-3} r!}
= 8 (-1)^{r-1} \binom{1/2}{r} \\
= -8\binom{1/2}{r}(-1)^r
$$
We know from the binomial theorem that $\sum_{r=0}^{\infty} \binom{n}{r} x^r = (1 + x)^n$ for $|x| < 1$, and with $x = -1$, we also know that for positive integer $n$, at least, we have $\sum_{r=0}^{\infty} \binom{n}{r} (-1)^r = (1-1)^n = 0$. In light of these, it is not hard to believe that $\sum_{r=0}^{\infty} \binom{1/2}{r} (-1)^r = 0$ as well: I'm not exactly sure how to prove this, but it would follow from a Tauberian theorem considering we can prove that the sum converges.
So our sum in this problem is
$$
1 + \sum_{r=3}^{\infty} -8 \binom{1/2}{r} (-1)^r 
= 1 - 8\sum_{r=3}^{\infty} \binom{1/2}{r} (-1)^r 
$$
The sum inside is almost the sum we said above is $0$, except that the $r = 0, 1, 2$ terms are missing. In other words, our sum in this problem is:
$$
\begin{align}
&1 - 8\Bigg(0 - \Big(1 + \frac12(-1) + \frac{(1/2)(-1/2)}{2}\Big)\Bigg) \\
&= 1 + 8\left(\frac38\right) \\
&= 4.
\end{align}
$$
