A friend of mine showed me this series and wanted me to find a $p(n)$ for this. Did it, but this series looks like something of a pre-established one. I'm not very much advanced into math, so does this series has some recognition?
I'll probably edit this question on how he derived this series or on what basis, but for the meantime this is what I've got.
So by the looks of it; (thanks to @PeterForeman)
$$p(n) = \frac{(n-1)!!}{n!!}$$
On
Actually, did the friend who showed you this imply that $p(n)$ was a polynomial? (That's what the symbol $p(n)$ usually means). If so neither Peter's nor my interpretation is a a polynomial. Coming up with a fourth degree polynomial where $p(x)= ax^4 + bx^3 + cx^2 + dx + e$ and $p(0)= e = 1$ and $p(1) =a+b+c+d+e=\frac 12$ and $p(2)= 16a+8b+4c+2d+e=\frac 23$ and $p(3)=81a+27b+9c+3d+e=\frac 38$ and $p(4)=256a+64b+16c+4d+e=\frac {8}{15}$ is fairly mechanical.
$e = 1$
$a+b+c+d+e=\frac 12$ so $a+b+c+d=-\frac 12$
$16a+8b+4c+2d+e=\frac 23$ so $16a+8b+4c+2d = -\frac 13$ so $8a+4b+2c+d=-\frac 16$ so $7a+3b+c=\frac 23$.
$81a+27b+9c+3d+e=\frac 38$ so $81a+27b+9c+3d=-\frac 58$ so $27a+9b+3c+d = -\frac 5{24}$ so $26a+8b+2c=\frac 7{24}$ so $13a+4b + c=\frac 7{48}$ and $6a+b=-\frac {25}{48}$.
And $256a+64b+16c+4d+e=\frac {8}{15}$ so $256a + 64b + 16c + 4d = -\frac 7{15}. So $64a + 16b + 4c +d = -\frac 7{60}$. So $63a+15b+3c=\frac {23}{30}$ and $(63-3*7)a + (15-3*3)b=42a+6b =\frac {23}{30} -2= -\frac {37}{30}$. So $7a + b=-\frac{37}{210}$.
So $a = -\frac{37}{210}+\frac {25}{48}=\frac {-37*8 + 25*35}{1680}=\frac {579}{1680}$
$b = -\frac{37}{210}-7*\frac {579}{1680} =-\frac {25}{48}-6*\frac {579}{1680}= -\frac{37}{210} -\frac{579}{240} = -\frac{25}{48}-\frac {579}{280}=-\frac{37*8}{210*8}-\frac{579*7}{1680}= -\frac{25*35}{1680}-\frac{579*6}{1680}=-\frac{4349}{1680}$.
$c = \frac 23 - 7*\frac {579}{1680}+ 3*\frac{4349}{1680}=\frac{1120}{1680}+\frac{8994}{1680}=\frac {10114}{1680}$.
And $d = -\frac 12 - a -b-c = \frac {-840 -579+4349-10114}{1680}= -\frac{7184}{1680}$.
So $p(n) = \frac {579}{1680}x^4 -\frac{4349}{1680}x^3 + \frac {10114}{1680}x^2 -\frac{7184}{1680}x + 1$ will give the same values.
In which case the next term is $\frac{579*625-4349*125 + 10114*25-7184*5}{1680} + 1 = \frac{35180}{1680} + 1= \frac {1843}{84}$.
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Peter Foreman says it looks like $a_n = \frac {n!!}{(n+1)!!}$ so that $a_5$ would be $\frac{15}{24}$. And $a_6$ would be $\frac {24}{75}$
I say it looks like $\frac{a_{2n}}{b_{2n}} = \frac {b_{2n-1}}{2*b_{2n-1}-1}$ and $\frac {a_{2n+1}}{b+{2n+1}} = \frac {b_{2n}}{2^{b_{2n}}}$ so that $a_5$ would be $\frac {15}{32,278}$. And $a_6$ would be $\frac {32,278}{65,535}$
Which one is correct? Which one is better?
Well, there is utterly no reason or method to prefer one over the other without more information. I'd say this question is unanswerable.
It looks like $$\begin{align} a_n &=\frac{n!!}{(n+1)!!}\\ &=\frac{n\cdot(n-2)\cdots(1\text{ or }2)}{(n+1)\cdot(n-1)\cdots(2\text{ or }1)}\\ \end{align}$$ where I use $n!!$ to denote the double factorial. But then again it could be an arbitrary polynomial without an infinite number terms.