1+4+10+20+35+...=?

Question

Is there a finite value to the infinite sum of all the tetrahedral numbers:

$$\sum_{n=1}^\infty \frac{n(n+1)(n+2)}{6}.$$

I know it's a divergent series, but I hear that $$ 1+2+3+4+5+\cdots=-\frac{1}{12}.$$ I'm wondering that $$1+4+10+20+35+\cdots$$ has the finite answer in the same sense.

Answer 1

The Riemann Zeta function is defined as $\zeta(s) = \displaystyle\sum_{n = 1}^{\infty}n^{-s}$, which is a convergent sum for $\Re(s) > 1$. However, the Riemann Zeta function does have an analytic extension to other values of $s$. Using this analytic extension, we have $\zeta(-1) = -\dfrac{1}{12}$. So in that sense, $\displaystyle\sum_{n = 1}^{\infty}n$ "equals" $-\dfrac{1}{12}$.

For your sum, we have $\displaystyle\sum_{n = 1}^{\infty}\dbinom{n+2}{3} = \displaystyle\sum_{n = 1}^{\infty}\dfrac{(n+2)(n+1)n}{6} = \displaystyle\sum_{n = 1}^{\infty}\dfrac{n^3+3n^2+2n}{6}$, which in some sense "equals" $\dfrac{\zeta(-3)+3\zeta(-2)+2\zeta(-1)}{6} = \dfrac{\tfrac{1}{120}+3 \cdot 0 + 2 \cdot -\tfrac{1}{12}}{6} = -\dfrac{19}{720}$.

Answer 2

The whole 'myth' that $\sum_1^\infty n = -\frac{1}{12}$ is just nonsense anyway.

You have to think about it like this:

You might know that $\frac{1}{1-z}=\sum{z^i}$ for $|z|<1$. So on the disk $|z|<1$ there is a connection between the value of $\frac{1}{1-z}$ and the value of $\sum{z^i}$. However, as soon as you move out of this disk of convergence that connection is completely lost; $\frac{1}{1-2}=-1$ while $\sum 2^i$ diverges.

Exactly the same thing happen with the $\sum_1^\infty n = -\frac{1}{12}$ myth: for $\Re (z)>1$ we have $\zeta(z)=\sum n^{-z}$. Note that is set $z=-1$ in the rhs side we obtain $\sum n^{1}=\sum n$. Now the whole myth stems from the fact that $\zeta(-1)=-\frac{1}{12}$. However! $\Re(-1)<1$, so there is no connection between the value of $\zeta(-1)$ and the value of $\sum n$!

Basically, whoever says $\sum_1^\infty n = -\frac{1}{12}$ should also believe that $-1=\infty$.

Answer 3

Here's an attempt using the generalized binomial theorem which doesn't work for this case but I find interesting anyway:

Let $f_m(x) =(1-x)^{-m} $.

For $|x| < 1$,

$\begin{array}\\ f_m(x) &=(1-x)^{-m}\\ &=\sum_{n=0}^{\infty} (-1)^nx^n \binom{-m}{n}\\ &=\sum_{n=0}^{\infty} (-1)^nx^n \frac{\prod_{k=0}^{n-1} (-m-k)}{n!}\\ &=\sum_{n=0}^{\infty} x^n \frac{\prod_{k=0}^{n-1} (m+k)}{n!}\\ &=\sum_{n=0}^{\infty} x^n \frac{(m+n-1)!/(m-1)!}{n!}\\ &=\sum_{n=0}^{\infty} x^n \binom{m+n-1}{m-1}\\ \end{array} $

For $x \ne 1$, this allows a value to be assigned to $f_m(x)$. Unfortunately, this does not work for $x = 1$.

(Letting $x \to 1^-$ might lead to zeta function territory, but I don't know enough to do this off the top of my head.)

Examples:

Setting $x = -1$, this gets $\sum_{n=0}^{\infty} (-1)^n \binom{m+n-1}{m-1} =\dfrac1{2^m} $. (For $m=1$, this is the usual $\sum_{n=0}^{\infty} (-1)^n =\dfrac1{2} $.)

Setting $x = 2$, this gets $\sum_{n=0}^{\infty} 2^n \binom{m+n-1}{m-1} =\dfrac1{(-1)^m} =(-1)^m $.

Answer 4

The below method works for the alternating sum of all the tetrahedral numbers:

Hint: Evaluate $~\displaystyle\sum_{n=1}^\infty(-x)^{n+2}~,~$ then differentiate it three times, divide it by six, and let $x=1$. See Abel summation for more information.

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Unfortunately, unlike in the case of the Riemann $\zeta$ and Dirichlet $\eta$ functions, here there is no “uniquely meaningful” way of relating the alternating and non-alternating versions of the same series to one another — which is not to say that various finite values cannot be ascribed to it, just that such values are not unique, as has already been pointed out in the comments to JimmyK’s answer $($which has been unjustly downvoted, by the way, since all the OP asked for was “a” value, not “the” value, and certainly no one can argue that the approach described there does not provide “one” such value, out of many possible others — but I digress$)$. In other words, in order for such a “special” value to exist, one must first find an alternate expression for $\displaystyle\sum_{n=1}^\infty\frac1{\displaystyle{n+2\choose3}^a}$ which would allow us to “generalize” the formula to values of $a\le\dfrac13$ , just as has been done in the case of the afore-mentioned $\zeta$ function.