Sylvester's Sequence is defined as $$e_{n+1}=e_n^2-e_n+1,\space\space e_0=2$$ and it has the interesting property that $$\sum_{k=0}^\infty \frac{1}{e_k}=1$$ despite the fact that there is no pretty closed-form explicit formula for $e_n$. This makes me curious about series of reciprocals of other sequences defined via quadratic recurrences, such as $$a_{n+1}=a_n^2+1,\space\space a_0=1$$ From this definition, I would like to find the value of $$\sum_{k=0}^\infty \frac{1}{a_k}$$ I have so far been unsuccessful in determining its value, and I would appreciate any ideas about how to do so. I have determined that to make the series telescope, it would suffice to find a function $f$ satisfying $$f(k)+\frac{1}{k}=f(k^2+1)$$ ... however, I have been unable to find such a function.
Any help is appreciated!
The reason the series of Sylvester sequence reciprocals works out so neatly is that $$ e_{n+1}-1=e_n(e_n-1)=\dots=\prod_{k=0}^{n}{e_k}\cdot(e_0-1)=\prod_{k=0}^{n}{e_k}, $$ so that $$ e_{n+1}=1+\prod_{k=0}^{n}{e_k}. $$ Then it is easy to see by induction that $$ 1-\sum_{k=0}^{n}{\frac{1}{e_k}}=\frac{1}{\prod_{k=0}^{n}{e_k}}=\frac{1}{e_{n+1}-1}, $$ since $$ \frac{1}{e_{n+1}-1}-\frac{1}{e_{n+1}}=\frac{1}{(e_{n+1}-1)e_{n+1}}=\frac{1}{e_{n+2}-1}. $$ Perhaps you can try to imitate what is going on here "for other values of $1$". I'm not sure, however, that the recurrence you suggest will work the same way. Perhaps, $a_{n+1}=a_n^2-ca_n+c$, $n\ge 1$, or something similar, and a suitably chosen $a_0$ will work out nicer. It won't be as pretty as the original, but it should work.