Evaluating product of exponent and polynomial In a probability theory problem, I need to solve an inequality over $n\in\mathbb{N}$ which can be expressed in a general form like this: 
$a^n ( b_1 n^2 + b_2 n + b_3) \leq c$ where $a, b_1, b_2, b_3, c$ are real numbers.
What is the standart way (or a good way) to tackle this? Is there an analytic solution?
For reference, here is the exact inequality:
$0.8^n ( \frac{1}{32} n^2 - \frac{7}{32} n + 1) \leq 0.1$
 A: There is no way in general to solve such equations (inequalities).
Since you have $a < 1$ there will be some $N$ such that all $n>N$ will satisfy your bound. One way to find $N$ is to guess powers of two until the statement is true. Then binary search until you find exactly $N$
viz:
$n=1 \rightarrow $ False 
$n=2 \rightarrow $ False 
$n=4 \rightarrow $ False 
$n=8 \rightarrow $ False 
$n=16 \rightarrow $ False 
$n=32 \rightarrow $ True
So $N \le 32$.
Then you'd binary search between 32 and 16 until you find that $N = 21$, for all $n\ge21 = N$ the inequality is satisfied.
If you had instead that $|a|>1$, then you'd need to find some negative value of $n$ to satisfy the inequality. If instead, $a = 1$, then you could actually solve it as a quadratic equation and find an analytic solution.  
A: Let's clean up a bit
$$ 0.8^n\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq 0.1 $$
$$ \left(\frac{8}{10}\right)^n\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq \frac{1}{10} $$
$$ \frac{8^n}{10^n}\left(\frac{1}{32}n^2-\frac{7}{32}n+1\right)\leq \frac{1}{10} $$
$$ \frac{8^n}{8}\left(\frac{1}{4}n^2-\frac{7}{4}n+8\right)\leq \frac{10^n}{10} $$
$$ 8^{n-1}\left(\frac{1}{4}n^2-\frac{7}{4}n+8\right)\leq 10^{n-1} $$
$$ \frac{1}{4}n^2-\frac{7}{4}n+8\leq \frac{10^{n-1}}{8^{n-1}} $$
$$ \frac{1}{4}\left(n^2-7n+32\right)\leq \left(\frac{5}{4}\right)^{n-1} $$
$$ n^2-7n+32\leq 4\left(\frac{5}{4}\right)^{n-1} $$
Now let's find the minimum of the quadratic function 
$$ \frac{d}{dn}\left[ n^2-7n+32\right]=2n-7=0 $$
$$ n=\frac{7}{2}=3.5 $$
However since $n\in\mathbb{N}$ then you can start with $n=4$ and work your way up from there. I would choose $n$ in increments of $10$ as it usually doesn't take long before an exponentially increasing function becomes greater than a quadratic function. After a few more than $3$ computations, you'll realize that $n\geq 21$ renders the inequality true.
