TITLE: Is it well known that for the Pythagorean spiral, $\sum_1^\infty\frac1{a_n+1} = \frac12$?
QUESTION [9 upvotes]: 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?

REPLY [4 votes]: 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.