Prove that for every positive integer $n$, $1/3 + 1/9 + \cdots + 1/{3^n} < 1/2$ Base case is $n=1$: $\frac {1}{3} < \frac{1}{2}$. So the base case holds.
Inductive hypothesis for $n = k$: $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} < \frac{1}{2}$
Inductive Step for $n = k + 1$:
$$ \left( \frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^k} \right) + \frac {1}{3^{k+1}}< \frac{1}{2}.$$
Multiplying the $n = k + 1$ step by $3$:
$$ \left( 1 + \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} \right) + \frac {1}{3^{k}}< \frac{3}{2}$$
$$\implies \frac{1}{3} + \cdots + \frac {1}{3^{k-1}} + \frac {1}{3^{k}} < \frac{3}{2} - 1 = \frac {1}{2}.$$
We know this to be true from our inductive hypoethsis. Hence, $\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} < \frac{1}{2}.$
Is this proof correct?

$$\sum\limits_{i=1}^n \frac{1}{3^i} = \frac{3n-1}{2 \cdot 3^n}$$
So $$\lim\limits_{n \to \infty} \frac{3n-1}{2 \cdot 3^n} = \frac{1}{2}$$
So for finite $k$:
$$\lim\limits_{n \to k} \sum\limits_{i=1}^n \frac{1}{3^i} < \frac{1}{2}$$
or for all finite $n$:
$$\sum\limits_{i=1}^n \frac{1}{3^i} + \underbrace{\sum\limits_{i=n+1}^\infty \frac{1}{3^i}}_{>0} = \frac{1}{2}$$
For $n \to \infty$, we get the equality.
Hence for all finite $n$ the bound holds.
