I've come across this problem in an advanced high school worksheet, but I don't know the origin of the problem.
Given the sequence $(x_n)_{\mathbb{N}}$
$$ \left\{ \begin{array}{l} x_{1}=1 \\ x_{n+1}=\sqrt{x_{n}(x_{n}+1)(x_{n}+2)(x_{n}+3)+1} , n\ge2 \end{array} \right. $$
How to evaluate the series
$$ \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{x_{k}+2} $$
Here's what I got:
$$ \begin{align} \sqrt{x_{n}(x_{n}+1)(x_{n}+2)(x_{n}+3)+1} & = \sqrt{(x_{n}^2+3x_{n}+1)^2} \\ & = |x_{n}^2+3x_{n}+1| \end{align} $$
As $x_1=1 > 0 $, $x_{n}^2+3x_{n}+1$ will be positive for all $n \ge 2$ . Thus:
$$ x_{n+1}=x_n^2+3x_{n}+1 $$
This is not good. The relation is neither linear nor homogeneous.
I also have no idea on how to evaluate the series. I can show that it converges comparing it to $\sum \frac{1}{k^2}$, but I can not think on how to evaluate it.
A quick program gives me these values for $x_n$ and for the partial sum $S_n$:
x[1]=1.0000000000
S[1]=0.3333333333
x[2]=5.0000000000
S[2]=0.4761904762
x[3]=41.0000000000
S[3]=0.4994462901
x[4]=1805.0000000000
S[4]=0.4999996936
x[5]=3263441.0000000000
S[5]=0.5000000000
x[6]=10650056950805.0000000000
S[6]=0.5000000000
x[7]=113423713055421845118910464.0000000000
S[7]=0.5000000000
x[8]=12864938683278672079501004830742670366487445279604736.0000000000
S[8]=0.5000000000
x[9]=165506647324519962593059552590935669575232079131271199714697991661245368701786157147328071653890281439232.0000000000
S[9]=0.5000000000
x[10]=27392450308603031764480870527381648752347551731785263407975564098191394906718645585260110781473383370045001182698658109358654445966317201271160135904061763917073783821325868369071377919640132022498004014465024.0000000000
S[10]=0.5000000000
So if I were to guess, I would say that $\sum \frac{1}{x_{n}+2} = 1/2$
I would appreciate any help finding a general expression for $x_n$ so I can go on to solving the series as usual.
After looking at the first few partial sums, it's easy to conjecture that the general form of the $n$th partial sum is $$ \sum_{k=1}^{n}{\frac{1}{x_k+2}}=\frac{1}{2}-\frac{1}{2}\prod_{k=1}^{n}{\frac{1}{x_k+2}}. $$ To show that, notice that $x_{n+1}+1=(x_n+1)(x_n+2)$, so $$ \frac{1}{x_n+2}=\frac{x_n+1}{x_{n+1}+1}, $$ so that $$ \frac{1}{2}\prod_{k=1}^{n}{\frac{1}{x_k+2}}=\frac{x_1+1}{2(x_{n+1}+1)}=\frac{1}{x_{n+1}+1}. $$ This makes the inductive step pretty easy: $$ \sum_{k=1}^{n+1}{\frac{1}{x_k+2}}=\left(\frac{1}{2}-\frac{1}{x_{n+1}+1}\right)+\frac{1}{x_{n+1}+2}=\frac{1}{2}-\frac{1}{(x_{n+1}+1)(x_{n+1}+2)}=\frac{1}{2}-\frac{1}{x_{n+2}+1}. $$
Update: Also, take a look at the OEIS page for A000058. These are the denominators $x_n+2$.