<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] The series can be rewritten as
$$
\begin{align}
&1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\dots\\
&8\left(\frac1{2\cdot4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+\dots\right)\\
&=8\sum_{k=1}^\infty\frac{(2k-1)!!}{(2k+2)!!}\tag{1}
\end{align}
$$
This is reminiscent of the series (obtained by the binomial theorem)
$$
\begin{align}
(1-x)^{-1/2}
&=1+\frac12x+\frac12\frac32\frac{x^2}{2!}+\frac12\frac32\frac52\frac{x^3}{3!}+\dots\\
&=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k)!!}x^k\tag{2}
\end{align}
$$
Substitute $x\mapsto x^2$ and multiply by $x$ to get
$$
x(1-x^2)^{-1/2}=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k)!!}x^{2k+1}\tag{3}
$$
Integration yields
$$
1-\sqrt{1-x^2}=\sum_{k=0}^\infty\frac{(2k-1)!!}{(2k+2)!!}x^{2k+2}\tag{4}
$$
Plugging in $x=1$ and subtracting $\frac12=\frac{(-1)!!}{2!!}$ yields
$$
\frac12=\sum_{k=1}^\infty\frac{(2k-1)!!}{(2k+2)!!}\tag{5}
$$
Multiplying by $8$ and applying $(1)$, we get
$$
\begin{align}
4
&=8\sum_{k=1}^\infty\frac{(2k-1)!!}{(2k+2)!!}\\
&=1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\dots\tag{6}
\end{align}
$$
</s>