<s>[INST] Find maximum of succession $a_{n+1}=\frac{n^2+n+4}{2n^2+1}a_n$ knowing $a_1=1$ I want to find out what's the maximum of the following sequence:
$$
\left\{ 
\begin{array}{c}
a_1=1 \\ 
a_{n+1}=\frac{n^2+n+4}{2n^2+1}a_n \\ 
\end{array}
\right. 
$$
I know that the limit of $a_n$ is equal to $0$. I have no clue about how to find the maximum. I have tried to find out the values of $a_1, a_2, a_3\dots$ but we couldn't find the maximum. Please note that $n\geq 0$. Any hints?
 [/INST] In general for a recursion of the form:
$$
a_{n+1} = f(n)a_{n}, a_{1} > 0.
$$
if $f(n) - 1$ has only one zero $> 0$ and $lim_{n\to\infty}f(n) < 1$ then you know $a_{n}$ will attain its maximum for the largest $n$ for which you have $f(n)\geq 1$.
in this case:
$f(n)= \frac{n^2+n+4}{2n^2+1} > 1\Rightarrow - n^2 +n +3 > 0 \Rightarrow n < \frac{1+\sqrt{13}}{2} \approx 2.303$. Thus your maximum is $a_{3}$ ...
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