Compute and find 2009th decimal(2009th digit after the point), without automation, the following sum

Question

Compute and find 2009th decimal of (2009th digit after the point), without automation, the following sum

$$\frac{10}{11}+\frac{10^2}{1221}+\frac{10^3}{123321}+ \cdots +\frac{10^9}{123456789987654321}$$

Answer 1

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 $9$.

Answer 2

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 $\dfrac{1}{\underbrace{99\ldots99}_{n}}=0.\overline{\underbrace{00\ldots00}_{n-1}1}$