Solving the inequality $1+1.2 + 1.2^2 + 1.2 ^3 + \cdots + 1.2^n < N/16?$

Question

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.

Answer 1

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 $$ 1 + \frac{n+1}{5} < 1 + \frac{N}{80} \implies n < \frac{N}{16} - 1. $$

Answer 2

$1+1.2+1.2^2+\ldots +1.2^n = \frac{1.2^{n+1}-1}{1.2-1}$ $= 5(1.2^{n+1}-1)$.

So you want to find the values of $n$ such that the inequality

$$5(1.2^{n+1}-1) < \frac{N}{16}$$

This gives

$$1.2^{n+1} < \frac{N}{80} - 1$$

Taking logs this gives

$$n \log 1.2 < \log \left(\frac{N}{80}-1\right)$$

which as $\log 1.2 < .2$ and $\log 80 \approx .4.4$ , will hold for large $N$ if the following is satisfied:

$$.2n < \log N - 4.5$$

or $$n < 5 \log N - 18$$