Sum of numbers 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 sum of digits of $S$ is $5+5=\boxed{10}$ .

S = 1 + 2 + 3 + . . . . . . . + 10^n ; S = [( 10^n ) / 2 ] + [(10^2n ) / 2 ] = [ 5 . 10^( n - 1 ) ] + [ 5 . 10^( 2n - 1 ) ] ; So for ( n = 1 ) ; S = 55 And for ( n = 2 ) ; S = 5050 ; for ( n = 3 ) ; S = 500500 ; And So On : So the Sum of All digits of ( S ) Is ( 5 + 5 = 10 ) : That Is True .

let $S = \sum\limits_{i=1}^{10^n-1} i$ then $9 | S$ , that implies the digits sum of $S$ must be 9. And the digits sum of $10^n$ is 1. So the answer is true.