Solving the inequality $1+1.2 + 1.2^2 + 1.2 ^3 + \cdots + 1.2^n < N/16?$ Given the inequality $$1+1.2 + 1.2^2 + 1.2 ^3 +\cdots + 1.2^n < \frac{N}{16}?$$
I need the value of $n$, or just an approximation. $N$ is known. 

Using the formula for the partial sum of a geometric series, rewrite this as
$$
\frac {1.2^{n+1}-1}{1.2-1} < \frac N{16} \implies
5(1.2^{n+1}-1) < \frac N{16} \implies
1.2^{n+1} < 1 + \frac N{80}.
$$
If we take a logarithm, we can solve this exactly to get
$$
n < \frac{\log(1 + \frac N{80})}{\log(1.2)} - 1.
$$
For a numerical approximation, we can note that $1.2^{n+1} \geq 1 + (0.2) \cdot (n+1)$, so that $n$ will necessarily satisfy the inequality if 
$$
1 + \frac{n+1}{5} < 1 + \frac{N}{80} \implies n < \frac{N}{16} - 1.
$$
