# Build general form of an infinite sequence

Please, I would like to build the general form of an infinite number sequence

$$\dfrac{8}{35}, \dfrac{5}{21} ,\dfrac{8}{33} ,\dfrac{35}{143},\dfrac{16}{65} ,\dfrac{21}{85} ,\dfrac{80}{323},\dfrac{33}{133},\dfrac{40}{161}, \cdots$$

as the form $$\dfrac{f(n)}{g(n)}$$. For example, I should get

$$\dfrac{8}{35}$$ for $$n =1$$ or $$n =2$$ (or $$n =3$$), $$\dfrac{5}{21}$$ for $$n =2$$, ....

I already asked this question on math.stackexchange.com but without success. I thought Mathematica could find this general form!

FindSequenceFunction[{8/35, 5/21, 8/33, 35/143, 16/65, 21/85,
80/323, 33/133, 40/161}, n] // FullSimplify

(*    ((1 + n) (3 + n))/((3 + 2 n) (5 + 2 n))    *)

• Great this Mathematica. Thank you @Roman, I'm extremely grateful. – Gallagher Mar 28 at 18:18

Using SequenceToSum:

 ResourceFunction["SequenceToSum"] [{8/35, 5/21, 8/33, 35/143, 16/65, 21/85, 80/323, 33/133,40/161, \[Ellipsis]}, n]

(*Inactive[Sum][(3 + 4 n + n^2)/(15 + 16 n + 4 n^2), {n, 1, \[Infinity]}]*)


$$\underset{n=1}{\overset{\infty }{\sum }}\frac{n^2+4 n+3}{4 n^2+16 n+15}$$

• This is the sum of the sequence. Since the sequence does not converge to zero (the sequence goes to 1/4), the sum diverges. – Bob Hanlon Mar 28 at 18:17
• Thank you so much @Mariusz Iwaniuk. I'm extremely grateful. But not the sum of termes, just the terms that I'm looking for. – Gallagher Mar 28 at 18:18
• @BobHanlon a bit of zeta function regularization can be used to make this sum converge: $\sum_{n=1}^{\infty} \frac{(n+1)(n+3)}{(2n+3)(2n+5)}=\sum_{n=1}^{\infty} \left(\frac{(n+1)(n+3)}{(2n+3)(2n+5)}-\frac14\right)+\sum_{n=1}^{\infty}\lim_{s\to0}\frac{n^s}{4}=-\frac{3}{40}+\frac14\lim_{s\to0}\zeta(-s)=-\frac{3}{40}-\frac18=-\frac15$. If you don't believe it, start reading here. – Roman Mar 28 at 19:40