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?
 A: As noticed you went in the wrong direction, for the induction step it is better to start from the induction hypothesis, for example as follows
$$\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} < \frac{1}{2}$$
$$\implies 1+\frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} <1+ \frac{1}{2}$$
$$\implies \frac{1}{3} + \frac{1}{9} + \cdots + \frac {1}{3^n} + \frac {1}{3^{n+1}} <  \frac{1}{3}+ \frac{1}{6}=  \frac{1}{2} $$
which complete the proof.
A: Consider a square of area $1$ and divide it into 3 congruent rectangles.
The area of each rectangle is $\frac{1}{3}$.
Select one of the three rectangles.
Divide one of the remaining two rectangles into three congruent rectangles, select only, as done before. Continue like this.
Irrespective of the number of iterations, the sum never reaches half of the original square.  
A: Consider, $$S=\frac{1}{3} + \frac{1}{9} + \frac{1}{27}+\cdots \tag{1}$$ Now multiply $S$ by 3, $$3S=1+\frac{1}{3} + \frac{1}{9} + \frac{1}{27}+\cdots \tag{2}$$ $(2)-(1)$, $$2S=1$$ $$S=\frac12$$
Thus, for finite terms, $$\frac{1}{3} + \frac{1}{9} + \frac{1}{27}+\cdots+\frac{1}{3^n}\lt\frac12$$
A: Also, we can use a formula for the sum of a geometric progression.
It's $$\frac{1}{3}\cdot\frac{1-\left(\frac{1}{3}\right)^{n}}{1-\frac{1}{3}}<\frac{1}{2}.$$
Can you end it now?
A: $$\sum\limits_{i=1}^n \frac{1}{3^i} = \frac{3n-1}{2 \cdot 3^n}$$
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.
