Closed form of $u_n$ with $u_{0} =1$ and $u_n=\frac{1}{3}u_{n-1}-1$? Let $u_n$ be a sequence defined by recurrence relation as : $u_{0} =1$ and $u_n=\frac{1}{3}u_{n-1}-1$, I'm not familiar to look for the closed form of $u_n $ , I have tried to find $u_n $ with a function of $n$ but I don't succeed because it's not defined as an arithmetic progression or a geometric progression. Then, is there any simple way to get it closed form ?
Thank for any help 
 A: I was about to suggest generating functions (for example), but characteristic polynomials will be easier and it's always good to have an arsenal of various tools.
$$3u_n=u_{n-1}-3 \iff 3=u_{n-1}-3u_n$$
same applies to
$$3=u_n-3u_{n+1}$$
altogether 
$$u_{n-1}-3u_n=u_n-3u_{n+1} \iff 3u_{n+1}-4u_n+u_{n-1}=0$$
with the following characteristic polynomial
$$3x^2-4x+1=0$$
with solutions $x_1=1$ and $x_2=\frac{1}{3}$, thus
$$u_n=Ax_1^n+Bx_2^n=A+B\frac{1}{3^n} \tag{1}$$
Since $u_0=1 \Rightarrow A+B=1$ and given $u_1=-\frac{2}{3} \Rightarrow -\frac{2}{3}=A+\frac{B}{3}$ we obtains $A=-\frac{3}{2}$ and $B=\frac{5}{2}$
and finally from $(1)$
$$u_n=-\frac{3}{2} + \frac{5}{2}\frac{1}{3^n}=\frac{5-3^{n+1}}{2\cdot 3^n}$$
A: You have to bring you back to a geometric progression, let say $v_n:=u_n+a$, with $v_n=\frac13 v_{n-1}$, $a$ to be determined.
$$v_n=\frac13 v_{n-1} \quad  \Rightarrow \quad u_n+a=\frac13 (u_{n-1}+a) \tag{1}$$
so (since $u_n=\frac13 u_{n-1} -1$ (2)), 
it comes ( (1)- (2))
$$ a = \frac13 a+1$$
so 
$$ a=\frac32 $$
$u_0=1$ so $v_0=\frac52$, so (since $v_n$ is a geometric progression)
$$ v_n =   \frac1{3^{n-1}}\frac52$$
and you can have $u_n$ :
$$ u_n=  \frac1{3^{n-1}}\frac52 -\frac32$$
A: You can express $u_n$ through $u_0$ sequentially through $u_{n-1}$, $u_{n-2}$ and so on
$$u_n =
\frac{1}{3} \cdot u_{n-1} - 1$$
Substitute $u_{n-1}$
$$u_n = \frac{1}{3} \left( \frac{1}{3} \cdot u_{n-2} - 1 \right) - 1 =
\frac{1}{3^2} \cdot u_{n-2} - \frac{1}{3} - 1$$
Then $u_{n-2}$ and so on
$$u_n =
\frac{1}{3^2} \left( \frac{1}{3} \cdot u_{n-3} - 1 \right) - \frac{1}{3} - 1 =
\frac{1}{3^3} \cdot u_{n-3} - \frac{1}{3^2} - \frac{1}{3} - 1 =
\dotsc$$
Finally
$$u_n = \frac{1}{3^n} - \sum \limits_{k = 1}^n \frac{1}{3^{n-k}}$$
To check this formula we can use mathematical induction.
The base case:
$$n = 1: u_1 = \frac{1}{3} - 1$$
The inductive step:
$$u_{n+1} = \frac{1}{3} \cdot u_n - 1 =
\frac{1}{3} \left( \frac{1}{3^n} - \sum \limits_{k=1}^n \frac{1}{3^{n-k}} \right) - 1 =
\frac{1}{3^{n+1}} - \sum \limits_{k=1}^n \frac{1}{3^{n+1-k}} - 1$$
We can substitute
$$1 = \frac{1}{3^{n+1 - k}}$$
where $k = n+1$, thus
$$u_{n+1} = \frac{1}{3^{n+1}} - \sum \limits_{k=1}^{n+1} \frac{1}{3^{n+1-k}}$$
So the formula is right
