Finding the limit of a sequence (through expansion) So I want to show that the sequence $a_n$ converges as $n \rightarrow \infty$. I want to find the value which it converges to. The sequence $a_n$ $(n=1,2,...) $ is defined by $ a_{n+1} = \alpha a_n + 1 $, where $ \alpha < 1$
I go about it in the following way, first by expansion:
$$
\begin{split}
a_{n}
 &= \alpha a_{n-1} + 1 \\
 &= \alpha (\alpha a_{n-2} + 1) + 1\\
 &= \alpha (\alpha (\alpha a_{n-3} + 1)) + 1 \\
 &= \ldots \\
 &= \alpha^{n-1} a_1 + \sum_{i=0}^{n-2} \alpha^i
\end{split}
$$
So when $n \to \infty $ I say that (because $ \alpha < 1$):
$ \alpha^{n-1} \rightarrow 0$, so 
$ \alpha^{n-1} a_1 \rightarrow 0$. (is this true?)
And then:
$$
\sum_{i=0}^{n-2} \alpha^i = \sum_{i=0}^{\infty} \alpha^i = \frac{1}{1-\alpha}
$$
So I conclude that $a_n$ converges to $ \frac{1}{1-\alpha} $
Is this the correct way of thinking about finding the limit of the sequence? The part I am particularly worried/unsure about is where I say $ \alpha^{n-1} a_1 \rightarrow 0$.
Is this correctly done or have I made an illegal step by using $a_1$?
 A: A different way to approach the problem is to guess you have an exponential form of some sort and a constant, so $a_n = Ar^n + B$, and plug it in, to get
$$
\begin{split}
a_n &= \alpha a_{n-1} + 1 \\
Ar^n + B &= \alpha \left(Ar^{n-1} + B\right) + 1\\
Ar^n + B &= \alpha Ar^{n-1} + B\alpha + 1
\end{split}
$$
which is true for all $n$, so $Ar^n = A\alpha r^{n-1}$ and $B = B\alpha +1$. The first equation implies $r = \alpha$ and the second $B = \frac{1}{1-\alpha}$, so we end up with
$$
a_n = A \alpha^n + \frac{1}{1-\alpha},
$$
and indeed if $\alpha^n \to 0$, we must have $a_n \to \frac{1}{1-\alpha}$.
A: I like to make sequences telescope.
If
$a_{n+1} = c a_n + 1
$
then,
dividing by $c^{n+1}$,
we get
$\dfrac{a_{n+1}}{c^{n+1}}
 = \dfrac{c a_n}{c^{n+1}} + \dfrac1{c^{n+1}}
=\dfrac{ a_n}{c^n} + \dfrac1{c^{n+1}}
$.
Letting
$b_n = \dfrac{ a_n}{c^n} 
$,
this becomes
$b_{n+1}
=b_n+\dfrac1{c^{n+1}}
$
or
$b_{n+1}-b_n
=\dfrac1{c^{n+1}}
$.
Summing
(can start at 0 or 1),
$\begin{array}\\
b_m-b_0
&=\sum_{n=0}^{m-1}(b_{n+1}-b_n)\\
&=\sum_{n=0}^{m-1}\dfrac1{c^{n+1}}\\
&=\dfrac1{c}\sum_{n=0}^{m-1}\dfrac1{c^{n}}\\
&=\dfrac1{c}\dfrac{1-\frac1{c^m}}{1-\frac1{c}}\\
&=\dfrac{1-\frac1{c^m}}{c-1}\\
\end{array}
$
or,
since
$b_0 = a_0$,
$\dfrac{ a_m}{c^m} 
=b_m
=a_0+\dfrac{1-\frac1{c^m}}{c-1}
$
so
$a_m
=a_0c^m+\dfrac{c^m-1}{c-1}
$.
Note that this works for
$c < 1$ and $c > 1$.
