Asymptote of largest root of polynomial of order $n$ I have a (log-) polynomial root equation given by
$$
f(n) = \log_2\left( \text{LargestRoot} \left[ 
\lambda^n - 3\sum_{i=0}^{n-1} \lambda^i = 0
\right]\right)
$$
The final equation of interest is then the fractional loss of $f(n)$ from 2:
$$
L(n) = \frac{2-f(n)}{2}
$$
Numerical solution for small $n$ is plotted below, and shows clear log-linear asymptotic behavior. The line shown is fit using $n \in [5, 20]$. I would like an analytical solution for the asymptote but don't see how to get there. Any help would be much appreciated.

 A: Let's first simplify the equation:
$$
\lambda^n-3\sum_{i=0}^{n-1}\lambda^i=\lambda^n-3\frac{1-\lambda^n}{1-\lambda}=0
$$
and thus
$$
\lambda^{n+1}-4\lambda^n+3=0
$$
(which has the same roots as the original equation, and an additional one at $\lambda=1$ because we multplied by $\lambda-1$). Since the largest root appears to be converging to $\lambda=4$, let's substitute $\lambda=4+\epsilon$:
\begin{eqnarray}
(4+\epsilon)^{n+1}-4(4+\epsilon)^n+3
&=&
4^{n+1}+4^n(n+1)\epsilon-4^{n+1}-4^nn\epsilon+3+O\left(\epsilon^2\right)
\\
&=&
4^n\epsilon+3+O\left(\epsilon^2\right)
\\
&=&
0\;.
\end{eqnarray}
Thus $\epsilon\approx-\frac3{4^n}$, which yields
\begin{eqnarray}
f(n)
&=&
\log_2(4+\epsilon)
\\
&=&
\log_24+\log_2\left(1+\frac\epsilon4\right)
\\
&\approx&
2+\frac\epsilon{4\ln2}
\\
&\approx&
2-\frac3{4^{n+1}\ln2}
\end{eqnarray}
and thus
\begin{eqnarray}
L(n)
&=&
\frac{2-f(n)}2
\\
&\approx&
\frac3{8\ln2}4^{-n}
\end{eqnarray}
and
\begin{eqnarray}
\log_{10}L(n)
&\approx&\log_{10}\left(\frac3{8\ln2}\right)-n\log_{10}4
\\
&\approx&
-0.27-0.60n\;,
\end{eqnarray}
in agreement with your fit result.
