Determine the greatest value of $n$ for which $b > a$ Let $a_n$ and $b_n$ be two recursive sequences so that:
$a_{n+1} = a_{n} + 2000$ where $a_{1} = 500$
and
$b_{n+1} = \frac{b_{n}}{1,001}$ where $b_{1} = 50000$
Determine the greatest value of $n$ for which $b > a$.
I have no idea of how to solve this. Perhaps the following is a start?
$a_{n+1} = a_{n} + 2000 \Rightarrow a_n = a_{n+1} - 2000$
$b_{n+1} = \frac{b_{n}}{1,001} \Rightarrow b_{n} = 1,001b_{n+1}$ 
And then you divide them and try to find a value for $n$ so that the fraction is greater than 1?
 A: You should find a closed form for both sequences. I can tell you that
$$b_n = 5000 \cdot 1.001^{n-1} \\
a_n = 500 + 2000 \cdot (n-1) = 2000 n - 1500$$
Now you can solve $a_n = b_n$ for $n\in\mathbb R$ (an extension of the sequences). The resulting $n_0$ must then be rounded down for an answer to the question.
A: Because $a$ increases and $b$ decreases (by a factor of 1001 each time, which is a lot!), it's quite easy to work it out by hand really quickly rather than finding a closed form for the sequences. Though it might be better to find a closed form if you are intending to generalize for the starting values, one needs to only verify:
$a_{1} = 500$, $a_{2} = 2500$, and $a_{3} = 4500$, while
$b_{1} = 50000$, $b_{2} = \frac{50000}{1001} < 2500$ (because $50000<2500000 <2500*1001$), and $b$ decreases from here while $a$ increases. Thus, $n=1$ is the greatest $n$.
A: I believe you can simply work out a few cases by hand in a table for the answer.
Otherwise, find a simple formula for each term of the two sequences in this way:
Notice that $a_1 = 500, a_2 = 2000 + a_1, a_3 = 2000 + a_2 = 2 \times 2000 + a_1, \cdots$, which tells you that $a_n = 2000(n-1) + 500 = 2000n - 1500$.
Similarly, $b_n = \frac{b_1}{1001^{n-1}}$.
Now $$b_n > a_n \\ \implies \frac{50000}{1001^{n-1}} > 2000n - 1500 \\
\implies \frac{100}{1001^{n-1}} > 5n - 3$$ and so on.
