Find the best constant $c < 0$ for $c a_n = a_{n-1} + 2 a_{n-2} + 3 a_{n-3} + ... + n a_0$ Let's define the following recurrence relation:
$$
\begin{align}
a_0 &= 1 \\
c a_n &= a_{n-1} + 2 a_{n-2} + 3 a_{n-3} + ... + n a_0
\end{align}
$$
Find the best constant $c < 0, \forall n \in \mathbb{N}$ which is farest from zero such that
$$
\begin{align}
a_n &> 0, \quad \text{if n is even} \\
a_n &< 0, \quad \text{if n is odd} \\
\end{align}
$$
My guessed answer would be $-\frac{1}{4}$. But how to prove it?
My attempt so far is to reduce the recurrence relation:
$$
\begin{align}
a_n &= \sum\limits_{k=1}^n \frac{k a_{n-k}}{c} = \frac{a_{n-1}}{c} + \sum\limits_{k=2}^n \frac{(k-1) a_{n-k}}{c} + \sum\limits_{k=2}^n \frac{a_{n-k}}{c} \\
&= \frac{c + 1}{c} a_{n-1} + \sum\limits_{k=2}^n \frac{a_{n-k}}{c} \\
&= \frac{c + 1}{c} a_{n-1} + \frac{a_{n-2}}{c} + \sum\limits_{k=3}^n \frac{a_{n-k}}{c} \\
&= \frac{c + 1}{c} a_{n-1} + \frac{a_{n-2}}{c} + a_{n-1} - \frac{c + 1}{c} a_{n-2} \\
&= \frac{2c + 1}{c} a_{n-1} - a_{n-2} \\
\end{align}
$$
If I set my guess $c = -\frac{1}{4}$ in this reduced recurrence relation then:
$$
\begin{align}
a_n = -2 a_{n-1} - a_{n-2}
\end{align}
$$
How to proceed? Or is there any other way to prove it?

From the recurrent relation $a_n = \frac{2c + 1}{c} a_{n-1} - a_{n-2}$ with $a_0 = 1$ and $a_1 = 1/c$, you can obtain an explicit formula for $a_n$:
$$a_n = p_1x_1^n+p_2x_2^n$$
with $x_1, x_2$ are the roots (can be non real) of the equation : $x^2 =  \frac{2c + 1}{c} x - 1$ and $p_1,p_2$ are the solution of the system of 2 linear equations:
$$p_1 + p_2 = a_0 = 1$$
$$p_1 x_1 + p_2 x_2 = a_1 = 1/c$$
After obtaining the analytic formula of $a_n$, the problem becomes easy. You just need to study the sign of $a_n$ in case $n$ is odd or even.
