Closed form of recursively defined sequence Let $a_0, a_1 \in \mathbb R$. I am given the sequence $(a_n)_{n\in \mathbb N}$ recursively defined by $$a_n = \frac{2}{5} a_{n-2} + \frac{3}{5} a_{n-1}$$
for $n \geq 2$. I want to find an explicit term describing $a_n$, I tried to figure it out by writing out the first 3-4 terms but I can't really find a pattern since the coefficients are not monotone. Or is there another, easier way to show that the sequence converges?
 A: Hint to see that the sequence converges:
Note that the term $a_{n+2}$ is a weighted mean of its two predecessors.
Hint to find the limit:
Try with $a_0=0$ and $a_1=1$. Then, scale and translate.
A: Given the recurrence
$$
a_{\,n}  = \frac{3}
{5}a_{\,n - 1}  + \frac{2}
{5}a_{\,n - 2} \quad \left| {\;a_{\,n\, < \,0}  = 0} \right.
$$
in my opinion, the best way to solve it is through the generating function.
So let's write it as:
$$
a_{\,n}  - \frac{3}
{5}a_{\,n - 1}  - \frac{2}
{5}a_{\,n - 2}  - \left[ {1 = n} \right]a_{\,1}  - \left[ {0 = n} \right]a_{\,0}  = 0\quad \left| {\;a_{\,n\, < \,0}  = 0} \right.
$$
where $[P]$ is the Iverson bracket
$$
\left[ P \right] = \left\{ {\begin{array}{*{20}c}
   1 & {P = TRUE}  \\
   0 & {P = FALSE}  \\
 \end{array} } \right.
$$
Then, multiplying by $z^n$ and summing
$$
\begin{gathered}
  0 = \sum\limits_{0\, \leqslant \,n} {a_{\,n} \;z^{\,n} }  - \frac{3}
{5}\sum\limits_{0\, \leqslant \,n} {a_{\,n - 1} z^{\,n} }  - \frac{2}
{5}\sum\limits_{0\, \leqslant \,n} {a_{\,n - 2} z^{\,n} }  - a_{\,1} \sum\limits_{0\, \leqslant \,n} {\left[ {1 = n} \right]z^{\,n} }  - a_{\,0} \sum\limits_{0\, \leqslant \,n} {\left[ {0 = n} \right]z^{\,n} }  =  \hfill \\
   = \sum\limits_{0\, \leqslant \,n} {a_{\,n} \;z^{\,n} }  - \frac{3}
{5}z\sum\limits_{0\, \leqslant \,n} {a_{\,n - 1} z^{\,n - 1} }  - \frac{2}
{5}z^{\,2} \sum\limits_{0\, \leqslant \,n} {a_{\,n - 2} z^{\,n} }  - a_{\,1} \;z - a_{\,0}  =  \hfill \\
   = A(z) - \frac{3}
{5}z\,A(z) - \frac{2}
{5}z^{\,2} \,A(z) - a_{\,1} \;z - a_{\,0}  \hfill \\ 
\end{gathered} 
$$
which gives:
$$
\begin{gathered}
  A(z) = \frac{{a_{\,1} \;z + a_{\,0} }}
{{1 - \frac{3}
{5}\,z - \frac{2}
{5}z^{\,2} }} = 5\frac{{a_{\,1} \;z + a_{\,0} }}
{{5 - 3\,z - 2z^{\,2} }} =  \hfill \\
   = \frac{5}
{7}\left( {\frac{{a_{\,1} \; + a_{\,0} }}
{{1 - z}} + \frac{{2a_{\,0}  - 5a_{\,1} }}
{{5 + 2z}}} \right) =  \hfill \\
   = \frac{1}
{7}\left( {\frac{{5\left( {a_{\,0}  + a_{\,1} } \right)}}
{{1 - z}} + \frac{{\left( {2a_{\,0}  - 5a_{\,1} } \right)}}
{{\left( {1 - \left( { - \frac{2}
{5}z} \right)} \right)}}} \right) =  \hfill \\
   = \frac{1}
{7}\left( {\frac{{5\left( {a_{\,0}  + a_{\,1} } \right)}}
{{1 - z}} + \frac{{\left( {2a_{\,0}  - 5a_{\,1} } \right)}}
{{\left( {1 - \left( { - \frac{2}
{5}z} \right)} \right)}}} \right) =  \hfill \\
   = \frac{1}
{7}\left( {\sum\limits_{0\, \leqslant \,n} {\left( {5\left( {a_{\,0}  + a_{\,1} } \right) + \left( {2a_{\,0}  - 5a_{\,1} } \right)\left( { - 1} \right)^{\,n} \frac{{2^{\,n} }}
{{5^{\,n} }}} \right)\;z^{\,n} } } \right) =  \hfill \\
   = \frac{{5\left( {a_{\,0}  + a_{\,1} } \right)}}
{7}\left( {\sum\limits_{0\, \leqslant \,n} {\left( {1 + \left( {\frac{{2a_{\,0}  - 5a_{\,1} }}
{{a_{\,0}  + a_{\,1} }}} \right)\left( { - 1} \right)^{\,n} \frac{{2^{\,n} }}
{{5^{\,n + 1} }}} \right)\;z^{\,n} } } \right) \hfill \\ 
\end{gathered} 
$$
That tells you that $a(n)$ is converging, oscillating, to $5/7(a_0+a_1)$.
