Limit of a sequence Calculate the limit of the sequence $\{x_n\},$ defined as follows: $$x_n=\dfrac{a+aa+aaa+aaaa\cdots+aaaaaaa..aaa }{10^n},$$ where $a\in\{1,2\ldots,9\}.$ 
$aaaaaaa..aaa = a , n$ time
Can anyone help ?   
 A: It looks like you mean the following:
$$x_n = \frac{a}{10^n}\sum_{k=1}^{n}\sum_{j=0}^{k-1}10^j$$
Assuming this is the case, first note that the inner sum can be computed in closed form. For any constant $c \neq 1$, we have
$$\sum_{j=0}^{k-1}c^j = \frac{c^k - 1}{c - 1}$$
so for $c = 10$ this is
$$\sum_{j=0}^{k-1}10^j = \frac{1}{9}(10^k - 1)$$
Substituting this back into the original expression gives us
$$x_n = \frac{a}{9\cdot 10^n}\sum_{k=1}^{n}(10^k - 1)$$
We can evaluate the sum as follows:
$$\begin{aligned}
\sum_{k=1}^{n}(10^k - 1)
&= \sum_{k=1}^{n}10^k - \sum_{k=1}^{n}1 \\
&= 10\sum_{k=0}^{n-1}10^k - n \\
&= \frac{10}{9}(10^n - 1) - n \\
\end{aligned}$$
Substituting this into the previous expression, we end up with
$$\begin{aligned}
x_n &= \frac{a}{9\cdot 10^n}\left( \frac{10}{9}(10^n - 1) - n \right) \\
&= \frac{10a}{81}\left( 1 - \frac{1}{10^n}  \right) - \frac{na}{9 \cdot 10^n}\\
\end{aligned}$$
In the limit as $n \to \infty$, the second term converges to zero, and the first term converges to $10a/81$. Therefore we conclude that
$$\lim_{n \to \infty}x_n = \frac{10a}{81}$$
A: By the Stolz-Cesaro theorem, one has
\begin{eqnarray}
\lim_{n\to\infty}x_n&=&\lim_{n\to\infty}\dfrac{a+aa+aaa+aaaa
\cdots+aaaaaaa..aaa }{10^n}\\
&=&\lim_{n\to\infty}\dfrac{aaaaaaa..aaa}{10^n-10^{n-1}}\\
&=&\lim_{n\to\infty}\dfrac{a\sum_{k=0}^{n-1}10^k}{10^n-10^{n-1}}\\
&=&\lim_{n\to\infty}\dfrac{a\frac{10^n-1}{9}}{9\cdot 10^{n-1}}\\
&=&\frac{10a}{81}
\end{eqnarray}
