<s>[INST] The sum of series $\frac14+\frac{1\cdot3}{4\cdot6}+\cdots$ The problem I have here today is the following;
$$\frac{1}{4}+\frac{1\cdot3}{4\cdot6}+\frac{1\cdot3\cdot5}{4\cdot6\cdot8}+\cdots$$ the problem is exactly phrased like this (I can't say that the $\infty$ sign is a bit unnecessary at the end),
My Attempts
We can generalize this sum by noticing that each time indices get greater the denominator and numerators are multiplied by $n+2$ for each $n$ either in numerator or denominator, we take $\dfrac{1}{4}$ out of the sum first, so the sum is equal to $$\dfrac{1}{4}+\sum_{k=4} \frac{(k-3)(k-1)}{k(k+3)}$$ then we open up the brackets and we get; ....
Then I was a little stuck here because when I opened up these brackets and try to get the partitions of the sum one of them was logical $\displaystyle\sum\frac{3}{k(k+3)}=\sum\frac{1}{k}-\frac{1}{k+3}$. I couldn't carry it out longer. What do you suggest?
Is this a problem way above elementary solutions?
 [/INST] The solution below is inspired by this other solution here.
Note that
$$\begin{align}
f(r)&=\frac {1\cdot 3\cdot 5\cdot\cdots \cdot(2r+1)}{4\cdot 6\cdot 8\cdot \cdots \cdot(2r+4)}\\ 
&=2\cdot \underbrace{\boxed{\frac {1\cdot 3\cdot 5\cdot\cdots \cdot(2r+1)}{2\cdot 4\cdot 6\cdot\cdots \cdot(2r+2)}}}_{A_r}\cdot \frac 1{2r+4}\\ 
&=2\ A_r\ \left(1-\frac {2r+3}{2r+4}\right)\\
&=2\left(A_r-A_{r+1}\right)\\
\frac 14+\frac {1\cdot 3}{4\cdot 6}+\frac {1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots&=\sum_{r=0}^\infty f(r)\\
&=2\sum_{r=0}^\infty A_r-A_{r+1}\\
&=2\left(A_0-\lim_{r\to\infty}A_{r+1}\right)\\
&=2\left(\frac 12-0\right)\\
&=\color{red}1\end{align}$$

See also this for the limit of $A_{r+1}$.
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