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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!

Somone has an idea please.

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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))    *)
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  • $\begingroup$ Great this Mathematica. Thank you @Roman, I'm extremely grateful. $\endgroup$ – Gallagher Mar 28 at 18:18
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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]}]*)

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

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    $\begingroup$ This is the sum of the sequence. Since the sequence does not converge to zero (the sequence goes to 1/4), the sum diverges. $\endgroup$ – Bob Hanlon Mar 28 at 18:17
  • $\begingroup$ Thank you so much @Mariusz Iwaniuk. I'm extremely grateful. But not the sum of termes, just the terms that I'm looking for. $\endgroup$ – Gallagher Mar 28 at 18:18
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    $\begingroup$ @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. $\endgroup$ – Roman Mar 28 at 19:40

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