Some things are quite easy

The series is in the form of $a_n = \sum_{m=1}^nm = \frac{n(n+1)}{2}$ , hence $a_{1999} = \frac{1999(2000)}{2} = 1999000$

While I find Kay Xspre solution amazing, I solved it while using a little cheat: used C++ program to count the sum.

What I noticed was that to find $1999^{th}$ you need I need to add all the numbers up to $1999$ in such way: $1+2+3+4+5...+1999$ .

Hence, you can write a program that would calculate such sum using any kind of loop. There's what I did with C++ (it would work pretty similarly with other programming languages as well):

#include <iostream>

using namespace std;

int main() {

    int sum = 0;
    for (int i = 1; i <= 1999; i++)
   {
        sum+= i;
    }
    cout << sum;

    return 0;
 }

While I find that solving without use of computer is better practice, solving with use of programming languages can help to understand different kind of programming concepts