Limit of a Sequence Defined Recursively Fix $r > 1$ and $a_1$ = 1 and define the sequence recursively as $a_{n+1}$ = $\frac{1}{r}$$(a_n + r + 1)$. Find the limit of the sequence.
 A: Suppose a limit $L$ exists, then $a_{n+1} \to L$ and $a_n\to L$, so
$$
L = \frac{1}{r}(L + r + 1)
$$
$$
\Rightarrow L = \frac{r + 1}{r-1}
$$
If $r=1$, then
$$
a_{n+1} = a_n + 2
$$
so the sequence does not converge since it is unbounded.
A: If the sequence has a limit $a$ then this limit must obey $a=\frac1r(a+r+1)$, so ....
A: Suppose that $\displaystyle a_n < \frac{r+1}{r-1}$ (which is certainly true for $a_1$).  Then $$ a_{n+1} < \frac{1}{r}(\frac{r+1}{r-1} + r + 1) = \frac{r^2+r}{r(r-1)} = \frac{r+1}{r-1},$$ and $r+1>a_n(r-1)$ so that $$a_{n+1} > \frac{1}{r}(a_n(r-1)+a_n) = a_n.$$  So by induction on $n$ this sequence is bounded above and increasing, and thus has a limit.  As the other answers have observed, it follows by taking limits that $$ L = \lim_{n\rightarrow\infty} a_n = \lim_{n\rightarrow\infty}\frac{1}{r}(a_n+r+1) = \frac{1}{r}(L+r+1)$$ and hence $$\displaystyle L = \frac{r+1}{r-1}.$$
A: Well, the other answers showed the easy way to find the limit, here is the difficult way, more over we want to know that if the limit really exist:
$ra_{n+1}=a_n+r+1\\
ra_{n+1}-a_n=r+1 \\
ra_{n+2}-a_{n+1}=r+1 \\
ra_{n+1}-a_n=ra_{n+2}-a_{n+1} \\
ra_{n+2}-(r+1)a_{n+1}+a_n=0 ,(r>1 \to \frac 1r \neq 1) \\ 
a_n=A1^n+B(1/r)^n \to \lim_{n\to\infty}a_n=A \\
a_0 = A+B \\
a_1=\frac{a_0+r+1}{r}=A+\frac{B}{r} \to a_0=Ar+B-r-1 \\
0=A(r-1)-(r+1)\\
A=\frac{r+1}{r-1}
$
It may be interesting to note that $A$ does not depend on the initial conditions ($a_0$) but $B$ does.
A: This can also be done using induction:
$$a_1=1\implies a_2=\dfrac{2+r}{r}\implies a_3=\dfrac{2+2r+r^2}{r^2}\implies a_4=\dfrac{2+2r+2r^2+r^3}{r^3}\implies\dots\dots$$
$$$$
$$\implies a_n=\dfrac{2+2r+2r^2+\dots+2r^{n-2}+r^{n-1}}{r^{n-1}}=1+\dfrac{2}{r}+\dfrac{2}{r^{2}}+\dfrac{2}{r^{3}}+\dots\dots+\dfrac{2}{r^{n-1}}$$
$$\implies a_n=1+2\sum_{i=1}^{n-1}\dfrac{1}{r^i}$$
$$\implies a_n\stackrel{n\to\infty}{\longrightarrow}1+2\dfrac{\dfrac{1}{r}}{1-\dfrac{1}{r}}=\dfrac{r+1}{r-1}$$
