Number of digits To Infinity :

Let the sum be $S$ . Then

$\begin{aligned} S & = 1 +2+3+\cdots +10^n \\ & = \frac {10^n(10^n+1)}2 \\ & = (5 \times 10^{n-1})(10^n+1) \\ & = 5 × 10^{2n-1} + 5 × 10^{n-1} \\ & = 5 \underbrace{0000 \cdots 0000}_{\text{number of 0s} = n-1} 5 \underbrace{0000 \cdots 0000}_{\text{number of 0s} = n-1} \end{aligned}$

Therefore the number of digits of $S$ is $1+n-1+1+n-1=\boxed{2n}$ .

*Let : S = 1 + 2 + 3 + . . . . . + 10^n : S = [ (10^n) / 2 ] + [ ( 10^2n) / 2 ] = [ 5 x 10^(n-1) ] + [ 5 x 10^( 2n - 1 ) ] : As we know that the number of
digits of ( 10^n ) = n + 1 : So the number of digits of [ 10^( 2n - 1 ) ] = 2n .