Find the $2000^{th}$ digit of the series $1234567891011121314\cdots $ Find the $2000^{th}$ digit of the series $1234567891011121314\cdots $
where $123456789\underbrace{1}_{10\text{th digit}}~~\underbrace{0}_{11\text{th digit}}~~\underbrace{1}_{12\text{th digit}}~~\underbrace{1}_{13\text{th digit}}~~\underbrace{1}_{14\text{th digit}}~~\underbrace{2}_{15\text{th digit}} \cdots $
and so on.
I really have no clue how to begin, as it is totally different from the other sequence and series problems  ,thanks .
 A: The sequence you appear to have given is the concatenation of all natural numbers. As others have noted, there is no guarantee that this is the case, but if it is, the following solution should work.
There are 9 single digit numbers, 90 with 2 digits, 900 with 3 digits, and 9000 with 4 digits, and, generally, $9(10)^m$ total numbers with $m$ digits.
$9+90\cdot2+900\cdot3=2889$, so you are in the 3 digit 'realm'. 
$9+90\cdot2+x\cdot3=2000$ gives $x=603.\overline{6}$, so you are looking for the $2^\text{nd}$ digit of the $604^\text{th}$ $3$ digit natural number. The first 3 digit number is $100$, therefor the $604^\text{th}$ 3 digit number is $100+603=703$. 
The last digit of that number would be represented by $x=604$, and $x=603.\overline{6}$ is the middle digit, and $x=603.\overline{3}$ would be the first digit. 
So the digit you are looking for is $0$. 
It is in the sequence here:
$$1\ 2\ 3\ 4\ 5\ldots \ 700\ \underset{604^\text{th}\text{ 3 digit natural number}}{701 \ 702\underbrace{703} 704\ 705\ 706\ldots} \ .$$

In general, the $N^\text{th}$ digit in that sequence can be found in the following way. Find the smallest $m$ such that 
$$N\leq\sum_{k=0}^{m}9k(10)^k.$$
If the above equation is true with equality, then you are looking for the last digit of the last natural number with $m$ digits. If the above equation an inequality, then solve for $x$:
$$N=mx+\sum_{k=0}^{m-1}9k(10)^k.$$
The solution will be
$$x=\left\lfloor x\right\rfloor+\frac{j}{m}.$$
And the digit you are looking for is the $j^\text{th}$ digit of the $\left\lceil x\right\rceil^\text{th}$ $m$ digit natural number.
A: This is not a mathematical answer, but just for the record, the Python code:
length=0
n=0
while length<2000:
    n+=1
    length+=len(str(n))
    print "\t", n, "\t", length

printet a list ending with
700     1992
701     1995
702     1998
703     2001

so the digit $3$ of the number $703$ is the $2001^{th}$ digit. Thus the $2000^{th}$ digit is confirmed to be the second digit of $703$, which is indeed $0$.
A: The above number is just a sequence of natural numbers.
Hence, from $1$ to $9$, there are $9$ numbers and $9$ digits; from $10$ to $99$, there are $90$ two-digit numbers and 180 digits (So, from 1 to $99$ there are $180 + 9 = 189$ digits) and from $100$ to $999$, there are $900$ three-digit numbers and $2700$ digits.
So, total number of digits from $1$ to $999$ = $9+180+2700 = 2889$
Starting from $1$, i.e. the $1^{st}$ digit (so $2$ is $2^{nd}$, $3$ is $3^{rd}$ and so on); we have to find the $2000^{th}$ digit of the given number
i.e. $2000-189 = 1811^{th}$ digit out of the three-digit numbers (Here I am removing all $1$-digit and $2$-digit numbers)
$1811 = (3)(603)+2$; which implies that our required digit is $2^{nd}$ digit of the $603^{rd}$ $3$-digit number, i.e. $2^{nd}$ digit of the $703^{rd}$ natural number from the starting, i.e. the $2^{nd}$ digit of $703$ i.e. $0$. Q.E.D.
