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

As stated in the first comment the easiest way is $$\sum_{n=1}^{\infty}n^2/2+n/2=\sum_{n=1}^{\infty}n^2/2+\sum_{n=1}^{\infty} n/2$$
Here's an alternative answer if you somehow do not want to use $\zeta(-1)$, nor the fact that $\zeta(-2)=0$ and don't want to do it by parts, by making it alternating, for d>1:
$$g(n)=n^2/2+n/2$$
$$\sum_{n=1}^{\infty}g(n)=\sum_{n=1}^{\infty}f(dn)-f(n)=\sum_{n=1}^{\infty}\sum_{p=1}^{d-1} f(n)e^{ip\pi2n/d}$$
$$f(n)=n^2c+nb$$
$$c=\frac{1}{(d^3-1)2}$$
$$b=\frac{1}{(d^2-1)2}$$
Say you want d=2,
$$f(n)=n^2/14+n/6$$
$$\sum_{n=1}^{\infty}(n^2/14+n/6)*(-1)^n$$
Which is ofcourse Cesaro/abel sumable or whatever trick you want to apply. 
