Sum of all triangle numbers

Does anyone know the sum of all triangle numbers? I.e 1+3+6+10+15+21... I've tried everything, but it might help you if I tell you one useful discovery I've made:

I know that the sum of alternating triangle numbers, 1-3+6-10... Is equal to 1/8 and that to change 1+3+6... Into 1-3+6... You would subtract 6+20+42+70... which is every other triangular number (not the hexagonals) multiplied by two.

1/8 plus this value is 1+3+6+10+...

A final note: I tried to split the triangle numbers into hexagonals and that series and then I got the squares of the odd numbers. Using dirichlet lambda functions This gave me 0 but I don't think this could be right. A number of other sums gave me -1/24 and 3/8 but I have no idea

Hint:

$$\sum_{k=1}^n\frac{k(k+1)}2=\frac12\sum_{k=1}^nk^2+\frac12\sum_{k=1}^nk$$

and the sum follows at once if you know

$$\sum_{k=1}^mk^2=\frac{n(n+1)(2n+1)}6$$

The sum of all the triangular numers, i.e. an infinite series, clearly diverges (and this, in this sense, the sum doesn't exist).

Let $g(n)$ denote the $n$th triangular number. Then $f(n)=\sum \limits _{i=1}^n g(n) = n(n+1)(n+2)/6$. Try proofing this via induction.