Find maximum of succession $a_{n+1}=\frac{n^2+n+4}{2n^2+1}a_n$ knowing $a_1=1$

Question

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?

Answer 1

It follows from your definition that $a_{n+1}>a_n$ is $n=1$ or $n=2$ and that $a_{n+1}<a_n$ otherwise. Therefore, the maximum is attained when $n=3$. That maximum is $\frac{20}9$, since $a_3=\frac{20}9$.

Answer 2

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 $a_{3}$ ...