Compute and find 2009th decimal(2009th digit after the point), without automation, the following sum Compute and find 2009th decimal of (2009th digit after the point), without automation, the following sum
$$\frac{10}{11}+\frac{10^2}{1221}+\frac{10^3}{123321}+ \cdots +\frac{10^9}{123456789987654321}$$
 A: You can write your series as
$$f(x)=\sum_{1}^{x}\frac{81\times10^k}{(10^k-1)(10^{k+1}-1)}=9\sum_{1}^{x}\frac{1}{10^k-1}-\frac{1}{10^{k+1}-1}$$
Where $x=9$. By telescopy we can show:
$$f(x)=\frac{10^{x+1}-10}{10^{x+1}-1}=1-\frac{9}{10^{x+1}-1}$$
So your sum is $S=1-1111111111^{-1}=0.\overline{9999999990}$. So since $2009\equiv 9\pmod{10}$, the digit is $9$.
A: Let
$$a_n=\sum\limits_{b=1}^n\frac{10^b}{12\ldots bb\ldots 21}$$
where $b$ is a digit less than or equal to $9$.
We prove inductively that $a_n=1-\dfrac{1}{\underbrace{11\ldots11}_{n+1}}$.
Base case: $\dfrac{10}{11}=1-\dfrac{1}{11}$
Now assume 
$$a_k=1-\dfrac{1}{\underbrace{11\ldots11}_{k+1}}$$
then 
$$\begin{align}
a_{k+1}&=1-\dfrac{1}{\underbrace{11\ldots11}_{k+1}}+\frac{10^{k+1}}{12\ldots (k+1)(k+1)\ldots 21}
\\&=1-\dfrac{1}{\underbrace{11\ldots11}_{k+1}}+\frac{10^{k+1}}{\underbrace{(11\ldots11)}_{k+1}~\underbrace{(11\ldots11)}_{k+2}}
\\&=1-\frac{\overbrace{(11\ldots11)}^{k+2}-10^{k+1}}{\underbrace{(11\ldots11)}_{k+1}~\underbrace{(11\ldots11)}_{k+2}}
\\&=1-\frac{\overbrace{(11\ldots11)}^{k+1}}{\underbrace{(11\ldots11)}_{k+1}~\underbrace{(11\ldots11)}_{k+2}}
\\&=1-\frac{1}{\underbrace{(11\ldots11)}_{k+2}}
\end{align}$$
as desired.
Now we just need to find the $2009$th decimal of 
$$a_9=1-\frac{1}{\underbrace{(11\ldots11)}_{10}}$$
but this is easy because this decimal is just $0.\overline{9999999990}$ $~~$ ($9$s everywhere except for every $10$th digit which is a $0$). Since $10\not|2009$, the digit we want is a $9$.
Note the easy decimal expansion comes from the fact that $\dfrac{1}{\underbrace{99\ldots99}_{n}}=0.\overline{\underbrace{00\ldots00}_{n-1}1}$
