Prove $\sum_{k=0}^n 3^k = O(3^n)$.
Below there is a picture from my text that contains the proof. My question pertains to the notation and/or assumptions in the proof. I don't need help with basic induction, below you'll see I'm not exactly sure what I'm confused about, it may be notation or possibly something quantifier related.
The author's explanation is giving me a bit of confusion near the end of the induction step when they require that
$$\left(\frac{1}{3} + \frac{1}{c}\right) \leq 1 \tag{A}$$
I'm not sure why the constant being less than $1$ is necessary. The way I was thinking of it (maybe my definition of big O notation is off) is that in our induction we should be verifying that
$$\text{if} \quad \sum_{k=0}^n{3^k} \leq c_1 3^n \quad \text{then} \quad \sum_{k=0}^{n+1}{3^k} \leq c_2 3^{n+1} \tag{B}$$
for some constants $c_1$ and $c_2$. If that's a valid induction approach then (by hijacking the author's proof)
$$\cdots \leq c_13^n + 3^{n+1} = \left(\frac{c_1}{3} + 1\right)3^{n+1}= c_23^{n+1} \tag{C}$$
Which would complete the proof.
T
Use the Geometric Series Theorem: $$\sum_{k=0}^n 3^k = {3^{n+1} - 1 \over 2} = (3/2)3^{n} - 1/2 = O(3^n)..$$