Is it well known that for the Pythagorean spiral, $\sum_1^\infty\frac1{a_n+1} = \frac12$?

Question

Let $\{a_n:n\in\Bbb Z^+\}$ be the sequence defined by $a_1=3$ and for $n\in\Bbb Z$ $$a_{n+1} = \frac12(a_n^2+1) $$ Note that all the $a_n$ are odd integers.

This is called the "Pythagorean Spiral," since each triplet of numbers $(a_n,a_{n+1}-1,a_{n+1})$ forms a pythagorean triplet:

$$ = a_n^2+ (a_{n+1}-1)^2 = a_{n+1}^2 $$

I have noticed (and can prove) that $$ \lim_{n\to\infty} \sum_{k=1}^n \frac1{a_k+1} = \frac12 $$

This is a fact dealing with well-traveled ground, and is not too difficult to prove (and easier still to notice). But I have not found this observation anywhere.

Is this a (relatively) well known property of the that for the Pythagorean spiral, and where could it be found?

Answer 1

Here is Sylvester's sequence whose reciprocals are known to sum to $1$. The sequence you have is Sylvester's sequence doubled minus $1$. So your sum of reciprocals summing to $1/2$ is equivalent to Sylvester's reciprocals summing to $1$.

Note how low the index is at OEIS for Sylvester's sequence. This indicates it's been studied a lot.

Answer 2

This is not an answer to Mark Fischler's question, but an approach for a proof that $\sum\limits_{n=1}^\infty\,\dfrac{1}{a_n+1}=\dfrac12$. The idea is to show that $$\frac{1}{a_n+1}=\frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}$$ for every $n=1,2,3,\ldots$. Then, by telescoping the sum, the claim follows.