<s>[INST] Determine value $\lim_{n\to +\infty} v_n$ Let $(u_n)$ sequence satisfy: $$\left\{\begin{matrix}u_1=3\\u_{n+1}=\frac{1}{5}\left(u_n^2+u_n+4\right),\: n=1, \: 2, ...\end{matrix}\right.$$
Set $v_n=\sum_{k=1}^n \frac{1}{u_k+3}$. Determine value $\lim_{n\to +\infty} v_n$
I think have to find a and b satisfy to $\frac{1}{u_k+3}=a\left(\frac{1}{u_k+b}-\frac{1}{u_{k+1}+b}\right)$
a and b=??
 [/INST] Let us expand 
$\large a(\frac{1}{u_k+b}-\frac{1}{u_{k+1}+b})$
Based on the given sequence recursion we obtain
$\large a(\frac{1}{u_k+b}-\frac{5}{u_k^2+u_k+4+5b})=a\frac{u_k^2+u_k+4+5b-5u_k-5b}{(u_k+b)(u_k^2+u_k+4+5b)}=a\frac{u_k^2-4u_k+4}{(u_k+b)(u_k^2+u_k+4+5b)}$
Simplifying further we have
$\large a\frac{(u_k-2)^2}{(u_k+b)(u_k^2+u_k+4+5b)}$
If you set $b$ to $-2$ the fraction becomes
$\large a\frac{(u_k-2)^2}{(u_k-2)(u_k^2+u_k-6)}=a\frac{(u_k-2)^2}{(u_k-2)^2(u_k+3)}=a\frac{1}{u_k+3}$
So $a$ is $1$.
Thus, $a=1$ and $b=-2$.  
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