Deriving closed-term formula from recurrence relation with linear algebra. I've seen an example on Wikipedia on how to derive Binet's by using eigenvectors and eigenvalues (here).
I tried to apply the same method to prove, for example, that $$\lim_{n\to\infty}\frac{1}{n^9}\sum_{k=0}^nk^8=\frac{1}{9}$$ (I know there are a bunch of other better methods to prove this, but I just want to learn something new)
The sequence can be written in a recursive form, such as
$$
a_{n+1}=\left(\frac{n-1}{n}\right)^9a_n+\frac{1}{n}
$$
Then I defined the vectorial recurring sequence
$$
\vec{a_{n+1}}=\begin{pmatrix}a_{n+1}\\1\end{pmatrix}=\begin{pmatrix}\left(\frac{n-1}{n}\right)^9&\frac{1}{n}\\0&1\end{pmatrix}\begin{pmatrix}a_{n}\\1\end{pmatrix}
$$
The eigenvalues of the matrix are clearly $\lambda_1=1$ and $\lambda_2=\left(\frac{n-1}{n}\right)^9$. The eigenvector relative to $\lambda_1$ is $\vec{\mu}_1=\begin{pmatrix}-\frac{n^8}{\left(n-1\right)^9-n^9}\\1\end{pmatrix}$
Then, since $a_0=1$,
$$
\vec{a_0}=\begin{pmatrix}1\\1\end{pmatrix}=\vec\mu_1
$$
So
$$
\vec{a_n}=A^n\vec\mu_1=\lambda_1^n\vec\mu_1=\vec\mu_1
$$
Therefore $$a_n=\frac{n^8}{(n-1)^9-n^9}$$
The limit we get from this is clearly $\frac{1}{9}$ as expected, but the nth term of this final sequence doesn't match with that in the limit.
What's wrong with my proof?

Making $b_n = (n-1)^9a_n$ you can solve the recurrence as
$$
b_{n+1}=b_n + n^8
$$
with solution
$$
b_n = c_0+\frac{n^9}{9}-\frac{n^8}{2}+\frac{2 n^7}{3}-\frac{7 n^5}{15}+\frac{2 n^3}{9}-\frac{n}{30}
$$
and then
$$
a_n = \frac{1}{(n-1)^9}\left(c_0+\frac{n^9}{9}-\frac{n^8}{2}+\frac{2 n^7}{3}-\frac{7 n^5}{15}+\frac{2 n^3}{9}-\frac{n}{30}\right)
$$
and clearly
$$
\lim_{n->\infty}a_n = \frac 19
$$
