<s>[INST] 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?
 [/INST] 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.
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