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
 A: 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).
A: 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.
A: Here is a start.
The sums of the triangle numbers come out as $$1,4,10,20,35,56$$
Differences between adjacent terms come out as $$3,6,10,15,21$$
Differences between these come out as $$3,4,5,6$$
And then the differences become constant $$1,1,1$$
Now you have to take differences three times to get a constant, so this means your formula will be a cubic in $n$. Also if you take differences $r$ times to get a non-zero constant $c$ then the coefficient of $n^r$ will be $\frac c{r!}$.
Here you expect a cubic which begins $\cfrac {n^3}6$
A: The $r$-th triangular number is 
$$T_r=\frac {r(r+1)}2=\binom {r+1}2$$
 i.e. $1, 3,6, 10, ...$ for $r=1, 2, 3, 4, ...$.
The sum of the first $n$ triangular numbers is
$$S_n=\sum_{r=1}^n T_r=\color{blue}{\sum_{r=1}^n \binom {r+1}2=\binom {n+2}3}=\frac {(n+2)(n+1)n}6$$
i.e. $1, 4, 10, 20, ...$ for $n=1, 2, 3, 4...$. This is also known as the $n$-th tetrahedral number. 
A: First: the given series is not Cesaro- or Abel-summable.      
For the Abel-summability we would make the ansatz 
$$ \lim_{x \to \;^-1} 1+3x+6x^2+\cdots+\binom{k+2}{2}x^k+ \cdots \underset{\mathcal A}= { 1\over(1-x)^3}\underset{\mathcal A}=+\infty
$$
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad $ $\mathcal A$ indicating Abel-summation for divergent series

Second: if one looks for alternatives, then shifting the index might come to mind, such that we might try
$$ \begin{array}{} &S_2&= &1+3+6+10+...  \\
 &&\underset{??}=& 1+2+3+4+... \\
 &&&+0+1+3+6+... \\
 &&\underset{??}=& S_1 + S_2 \qquad \qquad \qquad && &\text{where }S_1=1+2+3+4+... \\
 \implies& 0 &=& S_1 &&& \text{and no information for $S_2$}
\end{array}
$$
which makes such an idea worthless.          


Third: a more meaningful ansatz is surely the following. (This has already been proposed by Daniel Fischer, but I repeat this here to show the nice generalization)
By this we avoid the idea of any shifting of the index at all and formulate
$$ S_2(s) = {1\over 1^s} + {3\over 2^s} + {6\over 3^s} + \cdots +  {\binom{k+1}2\over k^s}+\cdots = \sum_{k=1}^\infty  {\binom{k+1}2\over k^s}
$$
which gives convergent series for all $s>3$ We can then observe (and prove, of course) that for the whole continuous range of convergence
$$ \begin{array}{}  2 S_2(s) &=& {2\over 1^s} + {6\over 2^s} + {12\over 3^s} + \cdots \\
 &=& {1\over 1^s} + {2\over 2^s} + {3\over 3^s} + \cdots \\
 &+& {1^1\over 1^s} + {2^2\over 2^s} + {3^2\over 3^s} + \cdots \\
 &=& \sum_{k=1}^\infty { 1\over k^{s-1}} + \sum_{k=1}^\infty { 1\over k^{s-2}} \\
 &=& 1 \cdot \zeta(s-1) +  1 \cdot \zeta(s-2) 
\end{array}$$
It must be taken from elsewhere, that we can extend this summation-expression beyond range of convergence, towards 
$$ S_2 = \lim_{s \to 0} \frac12 (1 \cdot \zeta(s-1)+ 1 \cdot \zeta(s-2))\underset{\mathcal Z}= \frac12 \cdot( - \frac1{12}+ 0) = -\frac1{24}
$$ 

I found it much interesting, how this could be generalized for series of binomials of higher order.
We get this way
$$ \begin{array}{}  
  S_1 &=& \sum_{k=1}^\infty \binom{k}{1} &\underset{\mathcal Z}=&1 \cdot \zeta(-1) \\
  S_2 &=& \sum_{k=1}^\infty \binom{k+1}{2}& \underset{\mathcal Z}=\frac1{2!} (&1 \cdot \zeta(-1) + 1 \cdot \zeta(-2) )\\
  S_3 &=& \sum_{k=1}^\infty \binom{k+2}{3} &\underset{\mathcal Z}=\frac1{3!} (&2 \cdot \zeta(-1) + 3 \cdot \zeta(-2) + 1 \cdot \zeta(-3) )\\
  S_4 &=& \sum_{k=1}^\infty \binom{k+3}{4} &\underset{\mathcal Z}=\frac1{4!} (&6 \cdot \zeta(-1) + 11 \cdot \zeta(-2) + 6 \cdot \zeta(-3) + 1 \cdot \zeta(-4))\\
\vdots & & \vdots
\end{array}$$
where your series occurs as $S_2$.                 

Heuristically I found the coefficients applying multiple regression (I can show this simple procedure if requested), and finally found, that those coefficients at the $\zeta(-k)$s are the unsigned Stirlingnumbers first kind.        
Displayed in a matrix (top-left from the infinite array):          


A: 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. 
