This is by far my most ambitious proof attempt to date and I'm not very good at them; so even if the proof is correct I would still appreciate feedback on nomenclature, clarity, elegance, etc...

Induction Hypothesis:

$\sum\limits_{k=1}^n \frac{1}{3^k} = \frac{3^n-1}{2\cdot 3^n}$

Show for $n+1$:

$\sum\limits_{k=1}^{n+1} \frac{1}{3^k} = \frac{1}{3} +...+\frac{1}{3^n} + \frac{1}{3^{n+1}} = \frac{3^n-1}{2 \cdot 3^n} + \frac{1}{3^{n+1}}$

$= \frac{3^{n+1}-3}{2 \cdot 3^{n+1}} + \frac{2}{2 \cdot 3^{n+1}}$

$= \frac{3^{n+1}-1}{2 \cdot 3^{n+1}}$

Base Case, $n=1$:

$\sum\limits_{k=1}^1 \frac{1}{3^k} = \frac{3-1}{2 \cdot 3} = \frac{1}{3}$


What this shows is that every partial sum of the series $\frac{1}{3} + \frac{1}{9} + ... + \frac{1}{3^n}$ has the form $\frac{3^n-1}{2 \cdot 3^n}$, so that $$\lim_{n \to \infty} \frac{3^n-1}{2 \cdot 3^n} = \frac{1}{2}$$

EDIT: At the advice of Thomas Andrews, I'm adding the proof of the limit:

$$\lim_{n \to \infty} \frac{3^n-1}{2 \cdot 3^n} = \lim_{n \to \infty} \frac{3^n}{2 \cdot 3^n} - \lim_{n \to \infty} \frac{1}{2 \cdot 3^n} = \frac{1}{2}$$

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    $\begingroup$ You ought to give a reason for that last conclusion about $$\lim_{n\to\infty} \frac{3^n-1}{2\cdot 3^n}=\frac12.$$ The rest looks okay. It might have been easier to write your induction hypothesis as $\sum = \frac{1}{2}-\frac{1}{2\cdot 3^n}$ - then the algebra might be a bit easier in the induction step. $\endgroup$ – Thomas Andrews Jun 9 '15 at 1:08
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    $\begingroup$ @Thomas Andrews $\lim_{n->\infty}(\frac{3^n-1}{2(3^n)})=\frac{1}{2} \lim_{n->\infty} (1-\frac{1}{3^n})=\frac{1}{2}$ $\endgroup$ – Aleksandar Jun 9 '15 at 1:09
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    $\begingroup$ @ThomasAndrews, do you mean something like "As $n$ grows to infinity, $-1$ becomes negligeable and therefore the expression becomes equivalent to $\frac{3^n}{2 \cdot 3^n}"$? If you mean something else, could you expand? $\endgroup$ – jeremy radcliff Jun 9 '15 at 1:14
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    $\begingroup$ That's just hand-wavy. How would you prove the limit? $\endgroup$ – Thomas Andrews Jun 9 '15 at 1:15
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    $\begingroup$ I think he might want to prove, $\sum_{k=0}^\infty (1/3^k)=1/2$ which is obviously not true. $\endgroup$ – Aleksandar Jun 9 '15 at 1:18

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