How to do this Math induction problem? Show that: 
$$\frac n3 + \frac n9 + \frac {n}{27} + \cdots = \frac n2.$$ 
When I start with $\frac 13 + \frac 19 + \frac {1}{27}$ it leads to a number close to $.5$ but it's not exactly $.5$.
 A: This is not (necessarily) an induction problem. Rather, it is a problem of real analysis, involving geometric series. It is a fact that for real $r$ with $-1<r<1,$ we have $$r^0+r^1+r^2+r^3+\cdots=\frac1{1-r},$$ and so $$r^1+r^2+r^3+\cdots=\frac1{1-r}-1.$$ In particular, for $r=\frac13,$ this becomes $$\frac13+\frac19+\frac1{27}+\cdots=\cfrac1{1-\frac13}-1=\cfrac1{\frac23}-1=\frac32-1=\frac12,$$ and so $$\frac{n}3+\frac{n}9+\frac{n}{27}+\cdots=n\cdot\left(\frac13+\frac19+\frac1{27}+\cdots\right)=n\cdot\frac12=\frac{n}2.$$

As an alternate approach, using sequences, you could let $a_1=\frac{n}3$ and $a_{k+1}=a_k+\frac{n}{3^{k+1}}$ for $k\ge 1$. Then you can show by induction (using the recursive definition) that $$\frac{n}2-a_k=\frac{n}{3^k\cdot 2}$$ for all $k.$ Hence, for large enough $k,$ we can make $\frac{n}2-a_k$ as small as we like, meaning $a_k\to \frac{n}2,$ or put another way, $$\frac{n}3+\frac{n}9+\frac{n}{27}+\cdots=\frac{n}2.$$
A: Neither of the above answers uses induction, and therefore neither is responsive to the question.  Hint to the OP:  Try an induction hypothesis of the form 
The sum of the first $k$ terms is 1/2-f(k)
where $f(k)$ is some function that clearly goes to zero.  
A: You have an infinite geometric series $\frac n3 + \frac n9 + \frac n{27}+\dots=n\sum_{i=1}^\infty\frac 1{3^n}$.  Using the terminology in the linked article, we have $a=\frac 13$ (the first term) and $r=\frac 13$ (the ratio between successive terms.  As long as $|r| \lt 1$ this infinite sum will converge to $\frac a{1-r}=\frac {\frac 13}{1-\frac 13}=\frac 12$
