<s>[INST] Finding the smallest value for $n$ such that $ \sum_{k=1}^{n}30\left(\frac{1}{3}\right)^{k-1} > 44.8 $ I had a test today where I was asked to find the smallest value for $n$ such that: 
$$ \sum_{k=1}^{n}30\left(\frac{1}{3}\right)^{k-1} > 44.8 $$ This question really frustrated me but this is how I went about:
This is a geometric sequence where $a = 30$ and $ r = \frac{1}{3} $ where $a$ is the first term and $r$ is the common ratio.
$$S_{n} = \frac{a(1-r^{n})}{1-r}$$
$$ \frac{30\left(1-\left(\frac{1}{3}\right)^{n}\right)}{1-\frac{1}{3}} > 44.8 $$
$$ 30\left(1-\left(\frac{1}{3}\right)^{n}\right) > 67.2 $$(Multiplied the right by $\frac{3}{2}$)
$$ -\left(\frac{1}{3}\right)^{n} > 1.24 $$
$$ \left(\frac{1}{3}\right)^{n} < -1.24 $$
But one cannot proceed as$ \left(\frac{1}{3}\right)^{n}$ cannot be negative. Also, I was told to use logs on both sides but one cannot have the log of a negative number. Where did I go wrong?
 [/INST] $\frac{30\left(1-\left(\frac{1}{3}\right)^{n}\right)}{1-\frac{1}{3}} > 44.8$
...this can be written as: $45(1-\frac{1}{3^n}) > 44.8$. And then:
$$
45(1-\frac{1}{3^n}) > 44.8  \\
-\frac{1}{3^n} > - \frac{1}{225} \\ 
225 < 3^n \qquad \text{for } n \geq 5
$$
</s>