$a_1 = 1, a_{n + 1} = 1 + \frac{1}{2}a_n.$ does this sequence converge? Consider the sequence that is defined recursively by $a_1 = 1, a_{n + 1} = 1 + \frac{1}{2}a_n.$ does this sequence converge and if so, to what? Prove it.
My answer is
since $a_2 = 1 + \frac{1}{2}, a_3 = 1 + \frac{1}{2} + \frac{1}{4}, a_4 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \dots$
Then yes the sequence converges and $a_n = \sum_{k=0}^{n-1} \frac{1}{2^k}$ and hence it is a finite geometric series and its sum is 2, am I correct?
 A: You could be a bit more rigorous about proving that
$$a_n=\sum_{k=0}^{n-1}\frac{1}{2^k}$$
using, for example, a simple induction argument, but other than that it looks fine. If you don't, for some reason, notice this pattern immediately, you could also show that the sequence in increasing and bounded above by, for example, $2$. Then it converges by a well known result. Then, to find the limit, simply take the limit of both sides of the recursion formula. If we denote the limit by $a$, then it becomes
$$a=1+\frac{1}{2}a\iff a=2.$$
The reason I show this method is because it is a more general approach that often works well for this type of problems, but your method of course works great here.
A: Your solution is ok but you can have a more general solution process if you recognise that this is a linear difference equation with constant coefficients.
The more general equation $a_{n+1}-\frac 12 a_n = f_n$ (in this particular case you have $f_n=1$ can be written as $a_n = a_{h,n} + a_{p,n}$, where $a_{h,n}$ is the general solution of the homogeneous equation $a_{n+1}-\frac 12 a_n = 0$ (in this particular case, $a_{h,n} = c (1/2)^n$) and a particular solution of the complete equation (in this case $a_{p,n} = 2$ is a constant particular solution). It should be noted that for more general $f_n$ it may be less trivial to guess the general expression for $a_n$ using your process.
The general solution is $a_n = c(1/2)^n +2$. The initial condition leads to $c = -2$, which means that $a_n= -(1/2)^{n-1} + 2$.
Now you see that the solution is a convergent sequence with $\lim a_n = 2$.
