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

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

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    $\begingroup$ One can obtain the bounds $0\leqslant n\leqslant 9$ for the $2009^{th}$ decimal digit. Is it good enough? $\endgroup$ – Pedro Tamaroff Jun 1 '14 at 3:08
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    $\begingroup$ (Maybe it is helpful)Those numbers are of the form $$1\times11,11\times111,111\times1111,\text{etc}$$ Why do you ask? $\endgroup$ – chubakueno Jun 1 '14 at 3:13
  • $\begingroup$ Just to note, it is not a problem I've made up, it's from RMT(revista de matematica timisoara), similar to the hungarian Komal. $\endgroup$ – shooting-squirrel Jun 1 '14 at 3:13
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    $\begingroup$ @chubakueno You're the 1...man...get it?...How did you see that? From here, the answer should be easy... make it your answer so I could vote it...if you can continue... $\endgroup$ – shooting-squirrel Jun 1 '14 at 3:17
  • $\begingroup$ They reminded me of the powers of $11$, and fooling around a little bit more got me that. Maybe setting them as $(10^k-1)/9$ will help in getting some congruence relation, I will try!. $\endgroup$ – chubakueno Jun 1 '14 at 3:19

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 $9$.

  • $\begingroup$ Is telescopy a fancy word for "looking at it and just kind of seeing a pattern"? Or does it actually mean something? $\endgroup$ – corsiKa Jun 1 '14 at 4:32
  • $\begingroup$ @corsiKa It is actually a thing :) $\endgroup$ – chubakueno Jun 1 '14 at 4:33
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    $\begingroup$ @corsiKa $\text{telescopy} \neq \text{telepathy}$ $\endgroup$ – wchargin Jun 1 '14 at 5:44

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

  • $\begingroup$ chupa made the initial crucial point, so I'm giving it to him... $\endgroup$ – shooting-squirrel Jun 1 '14 at 3:42
  • $\begingroup$ @shooting-squirrel Well, I did notice that before he mentioned it...I was just busy writing my Latex. However, it's true that his is a more elegant solution because writing everything in terms of $10^k$ makes it all more compact. $\endgroup$ – Peter Woolfitt Jun 1 '14 at 3:46

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