Repeating Decimal Proof I need to prove the sequence converges, and determine the limit
{$x_n$}, where $x_n$ = 0.142857142857142857...142857
I assume it converges to 1/7. But I'm kinda lost at how to attack this.
Thanks.
 A: We will end up with two different arguments. The first that uses facts about geometric series. The second, in the Remark towards the end, uses general facts about series.
Let $a=\dfrac{142857}{10^6}$. Your number is equal to 
$$a+ar+ar^2+ar^3+\cdots+ar^n+\cdots,\tag{$1$}$$
where $r=10^{-6}$. 
Now use standard facts about infinite geometric series.
If you know already that the  series $(1)$ converges to $\dfrac{a}{1-r}$, then only a little arithmetic is left to do. It all comes down to the fact that $7\times 142857=10^6-1$.  
If you need to prove convergence, use the formula for the sum of a finite geometric series to find the sum $S_n$ of the first $n$ terms, and show that $\lim_{n\to\infty}S_n=\frac{a}{1-r}$. 
The relevant formula for the sum $S_n$ of the finite geometric series 
$a+ar+ar^2+\cdots+ar^n$
is (when $r\ne 1$) given by
$$S_n=a\frac{1-r^{n+1}}{1-r}.$$
This can be proved by induction, or more casually by multiplying out $(a+ar+\cdots+ar^n)(1-r)$ and observing the mass cancellation.
Now if $|r|\lt 1$, then $r^{n+1}\to 0$ as $n\to \infty$, so the sum of the infinite geometric series $(1)$ is $\dfrac{a}{1-r}$. 
Remark: Maybe we should write a little more, as suggested by marty cohen, and consider first the series 
$$\frac{1}{10}+\frac{4}{10^2}+\frac{2}{10^3}+\frac{8}{10^4}+\frac{5}{10^5}+\frac{7}{10^6}+\frac{1}{10^7}+\frac{4}{2^8}+\cdots.$$
Then we can gather together terms as in the main post. For the argument about the limit, in effect we examined the partial sums of the series above only for $n$ a multiple of $6$. In principle we should examine partial sums for all $n$, or find another argument for convergence. 
If you have standard comparison tests already available, it is obvious that the partial sums of the series above are increasing and bounded. So the series does converge.  
That gives us another way of finding the sum. Let $T_n$ be the partial sum of the first $n$ terms of the above series. Since the sequence $(T_n)$ converges, so do the sequences $(T_{6k})$, $(T_{6k+6})$, and to the same limit $b$. We have
$$T_{6k+6}=\frac{142857}{10^6}+\frac{T_{6k}}{10^6}.$$
Taking limits we get that
$$b=\frac{142857}{10^6}+\frac{b}{10^6}.$$
Solve for $b$: we get, after some simplification, $\dfrac{1}{7}$. 
