<s>[INST] 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
$$\frac{10}{11}+\frac{10^2}{1221}+\frac{10^3}{123321}+ \cdots +\frac{10^9}{123456789987654321}$$
 [/INST] 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 $9$.
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