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) $4$.
 A: 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:
$$
\begin{align}
&1 - 8\Bigg(0 - \Big(1 + \frac12(-1) + \frac{(1/2)(-1/2)}{2}\Big)\Bigg) \\
&= 1 + 8\left(\frac38\right) \\
&= 4.
\end{align}
$$
A: 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
$$
\begin{align}
4
&=8\sum_{k=1}^\infty\frac{(2k-1)!!}{(2k+2)!!}\\
&=1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\dots\tag{6}
\end{align}
$$
A: Well , imagination is the magic and you will see the best and most elegent method that follows as such:
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots=1+\frac{1\cdot3\cdot(6-5)}{6}+\frac{1\cdot3\cdot5\cdot(8-7)}{6\cdot8}+\cdots$$
$$=1+\frac{1\cdot3\cdot6}{6}-\frac{1\cdot3\cdot5}{6}+\frac{1\cdot3\cdot5\cdot8}{6\cdot8}-\frac{1\cdot3\cdot5\cdot7}{6\cdot8}\cdots=1+1\cdot3=1+3=4$$
Well , this telescoping method really helps in majority of cases.:)
A: 
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
$$
\sum_{n=0}^{N-1}\prod_{k=1}^n\frac{2k+1}{2k+4}=4-4\cdot\frac1{4^N}{2N+1\choose N}
$$
A: As suggested in another answer, one can write:
$$\prod_{k=1}^n \frac{2k+1}{2k+4}=\frac{8}{\pi}\cdot \frac{\Gamma (n+\frac{3}{2})\Gamma (\frac{3}{2})}{\Gamma (n+3)}=\frac{8}{\pi}\text{B}\left(n+\frac{3}{2},\frac{3}{2}\right)$$
Now, using the definition of the beta function, our sum is:
$$1+\frac{8}{\pi}\sum_{n\geq 1}\int_0^1 t^{n+\frac{1}{2}}(1-t)^{\frac{1}{2}}\,dt=1+\frac{8}{\pi}\int_0^1 t^{\frac{3}{2}}(1-t)^{-\frac{1}{2}}\,dt$$
Letting $t=\sin^2 w,$ this gives
$$1+\frac{16}{\pi}\int_0^{\frac{\pi}{2}}\sin^4 w\,dw=1+\frac{16}{\pi}\left(\frac{3\pi}{16}\right)=4$$
A: I believe the general term after the first is
$$\frac{(2n+1)! \cdot 4 \cdot 2}{2^n \cdot n! \cdot 2^{n+2} \cdot (n+2)! } \ , $$
with $ \ n \ \ge \ 1 . $
The series then ought to be convergent.  The series is clearly larger than
$$1 \ + \  \sum_{n=0}^{\infty} \frac{1}{2}  \cdot \left( \ \frac{5}{8} \ \right)^n \ , $$
so the sum is bigger than 2, but not infinite.  That leaves 4 among the choices.
A: 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 $4\sum_{n=1}^{\infty} P(X=n) = 4$.
A: i suppose $a_n= \frac{1 \cdot 3 \cdot 5 \cdots (2n+1)}{6 \cdot 8 \cdots (4+2n)},\, n\ge1$.
Using Stirling's formula we can see $a_n \sim n^{-3/2}$. Then the series converges. For the other side, the series is bigger than  2.
