finding recursive formula and show it converges to a limit Suppose we are playing cards and we start with $1000$ dollars. Every hour we lose $\frac{1}{2}$ of our money and then we buy another $100$ dollars. I am trying to find $x_n$ for the amount of money the player has after $n$ hours.
I think we can just take $x_n = \frac{x_{n-1}}{2} + 100 $
An so, let $L = \lim x_n$. Then $L = \frac{L}{2} + 100 $ and so $L = 200$
Is this correct?
 A: What you did is not wrong, but it's not complete.
What you did prove:

If the sequence $x_n$ has a limit, then the limit is equal to $200$.

What you did not prove:

The sequence $x_n$ has a limit.


Also, that's not what the question is asking you. The question says you need to find a formula for $x_n$, not the limit of $x_n$.
A: Not quite right, you have to prove that the sequence has a limit first.
In fact, the value "200" is the fixed point of the sequence. sometimes the sequence converges, sometimes not. If the sequence converges, then the limit is equal to the value of the fixed point.
To figure out whether the sequence has a limit, you can draw two curves on the $x-y$ plane, the first is $y = x$, the second is $y = f(x)$ where $f(x)$ is the recurrence relation of the sequence.
find the point $(x_0, f(x_0))$, draw a horizontal line which meets $y = x$ at $(f(x_0),f(x_0))$, then draw a vertical line which meets $y = f(x)$ at $(f(x_0),f(f(x_0)))$...
After repeat the operation above, you will infinite amount of points on the line $y = f(x)$, if the points converges to the fixed point, then there is a limit of the sequence. 
A: $x_n = \frac{x_{n-1}}{2} + 100 = (\frac{\frac{x_{n-2}}{2} + 100}{2}) + 100 =\frac{x_{n-2}}{4} + 100.(\frac{1}{2} + 1)= \frac{\frac{x_{n-3}}{2} + 100}{4} + 100.(\frac{1}{2} + 1)=...$
$\implies x_n = \frac{x_0}{2^n} + 100.(\frac{1}{2^{n-1}} + \frac{1}{2^{n-2}} + ... + \frac{1}{2} + 1 ) $
where $x_0 = 1000$.
