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 sum (also an infinite sum) of integers is never a fraction between $0$ and $1$. – Peter Nov 23 '14 at 13:02
• If you take $$\sum_{n=1}^\infty \frac{T_n}{n^s},$$ you obtain $$\frac{1}{2}\left(\zeta(s-2) + \zeta(s-1)\right),$$ and plugging in $s = 0$ gives you $\frac{1}{2}\zeta(-1) = -\frac{1}{24}$. – Daniel Fischer Nov 23 '14 at 13:18
• Thanks! I ruled out zeta functions because I wasn't sure how I could obtain a series of triangular numbers. I didn't think of plugging zeta -1 into the triangular function – user194821 Nov 23 '14 at 13:28

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).

• Yeah, I was looking for infinite sums so I tried Ramanujan summation by multiplying by 4 and subtracting – user194821 Nov 23 '14 at 13:12
• @user194821, one necessary condition for any infinite series $\;\sum a_n\;$ to converge (exist) is that $\;\lim a_n=0\;$ . Here this doesn't happen. – Timbuc Nov 23 '14 at 13:15
• Note that the question is tagged divergent-series. The OP wants a value for $$\sum_{n=1}^\infty \frac{n(n+1)}{2}$$ using a summation method for divergent series. – Daniel Fischer Nov 23 '14 at 13:21
• Why is it obviously false? If you play the game of summing divergent series at all, then $$f(z) = \frac{1}{(1-z)^3} = \frac{1}{2}\left(\frac{d}{dz}\right)^2\frac{1}{1-z} = \frac{1}{2}\left(\frac{d}{dz}\right)^2\sum_{k=0}^\infty z^k = \sum_{n=1}^\infty \frac{n(n+1)}{2}z^{n-1}$$ gives you $$\sum_{n=1}^\infty (-1)^{n-1}T_n = f(-1) = \frac{1}{8}.$$ – Daniel Fischer Nov 23 '14 at 13:32
• @Timbuc I think the tag divergent-series makes it pretty clear what the OP wants. I don't like that game too much either, but it exists. – Daniel Fischer Nov 23 '14 at 13:37

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.

• Note that the question is tagged divergent-series. The OP wants a value for $$\sum_{n=1}^\infty \frac{n(n+1)}{2}$$ using a summation method for divergent series. – Daniel Fischer Nov 23 '14 at 13:20

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 $\cfrac {n^3}6$

• The first six are tetrahedral numbers and I noticed that from the partial sums of 1+2+3+4+5+... Are triangle numbers and the final sum is -1/12 then -1/12 is kinda like the last triangle number. The last tetrahedral number is the answer to the sum in question – user194821 Nov 23 '14 at 13:15
• Note that the question is tagged divergent-series. The OP wants a value for $$\sum_{n=1}^\infty \frac{n(n+1)}{2}$$ using a summation method for divergent series. – Daniel Fischer Nov 23 '14 at 13:21
• @DanielFischer Sorry I didn't see the tag – Mark Bennet Nov 23 '14 at 13:30

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.