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.

I think I may be misunderstanding what you are asking. As far as I am aware the fourth row of Pascal's triangle is finite, and its sum is $2^4$. — Joel, Jul 20 '15 at 16:44
@Joel: Presumably what is meant is $$\binom{3}{3}+\binom{4}{3}+\binom{5}{3}+\binom{6}{3}+\cdots$$ (the "diagonal" being a row if one draws the triangle unusually). — Zev Chonoles, Jul 20 '15 at 16:46
I think I see what you mean now, I would say this is the fourth diagonal of Pascal's triangle. What you are seeking is a resolution to the the divergent series: $$\sum_{n=3}^\infty { n \choose n-3 }.$$ — Joel, Jul 20 '15 at 16:48
Who made the 5 upvotes for the first comment ? The comment is: " I think I may be misunderstanding what you are asking. As far as I am aware the fourth row of Pascal's triangle is finite, and its sum is $2^4$". — callculus42, Jul 20 '15 at 16:48
@calculus nothing wrong with giving an upvote. This is clearly not a homework question, but a question to extend the OP knowledge of divergent series. The OP has indicated that they believe the resolution of this divergent series can be found via a Ramanujan's sum, but wants more information as to how to go about it or if it is even meaningful. — Joel, Jul 20 '15 at 16:51
Right Joel. I avoided using the limit infinity sign because I was afraid people will just think this is a divergent series. My mistake. — new, Jul 20 '15 at 16:52
@user251257 not a joke, you can assign a meaningful value to some diverging series. It was discovered by euler, and further developed by Riemann. — john, Jul 20 '15 at 16:52
Whoever is promoting the idea that that $1+2+3+4+5+\cdots=-\frac1{12}$ should answer this question. — John Bentin, Jul 20 '15 at 16:55
@john: oh right, I forgot the zeta function ... for many reasons. Thx. — user251257, Jul 20 '15 at 16:57
@Joel Sorry. It seems, that I have misunderstood the question. — callculus42, Jul 20 '15 at 16:58
The question is awful. If what is asked is the number Ramanujan's summation yields in this case, then the result is $$\frac16\left(-\frac{B_4}4-3\frac{B_3}3-2\frac{B_2}2\right)=\frac16\left(\frac{1}{30\cdot4}-0-\frac1{6}\right)=-\frac{19}{720}$$ — Did, Jul 20 '15 at 17:03
To be quite explicit, the main value of my previous comment is to provide a relevant link (certainly not the computation itself, which is boring at best). — Did, Jul 20 '15 at 17:06
^Based on the original wording of the question, my guess is that the computation is what the OP was looking for. I could be wrong though. — JimmyK4542, Jul 20 '15 at 17:51

JimmyK4542 · Answer 1 · 2015-07-20T17:53:23.007

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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" $\dfrac{\zeta(-3)+3\zeta(-2)+2\zeta(-1)}{6} = \dfrac{\tfrac{1}{120}+3 \cdot 0 + 2 \cdot -\tfrac{1}{12}}{6} = -\dfrac{19}{720}$.

edited Jul 20 '15 at 17:53

answered Jul 20 '15 at 16:58

JimmyK4542

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So basically I can divide and conquer infinity limit even if it comes to the divergent series? – new Jul 20 '15 at 17:01
This is the same that making the Ramanujan summation? – Masacroso Jul 20 '15 at 17:03
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Not to nitpick, but this manipulation really is not quite the same; I'm not an expert in complex analysis, but it seems to me you need to show that the function $$\frac{\zeta(s-3)+3\zeta(s-2)+2\zeta(s-1)}{6}$$ has an analytic continuation at $s=0$. Perhaps that follows easily from the analytic continuation of $\zeta(s)$, but I don't know. – Zev Chonoles Jul 20 '15 at 17:03
^I'm also not an expert in complex analysis, but yes, this answer isn't very rigorous. – JimmyK4542 Jul 20 '15 at 17:09
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The sum can equally well be written $$\frac16\sum_{n=1}^\infty[(n+1)^3-2(n+1)+1]=\frac16\sum_{n=1}^\infty(n^3-2n+1),$$which by your logic is $\frac16[\zeta(-3)-2\zeta(-1)+\zeta(0)]=-\frac{13}{240}.$ – John Bentin Jul 20 '15 at 17:27
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To expand on @John's comment, the divergent series that "equal" $\zeta(-n)$ are not linear nor stable, so they (and their compositions) cannot be regularized with this kind of argument. – Vincenzo Oliva Jul 20 '15 at 17:30
@John: Out of curiosity, how did you write the sum $\displaystyle\sum_{n = 1}^{\infty}\binom{n+2}{3}$ as $\displaystyle\dfrac{1}{6}\sum_{n = 1}^{\infty}[(n+1)^3-2(n+1)+1]$? – JimmyK4542 Jul 20 '15 at 17:44
@JimmyK4542: $n^3+3n^2+n=(n+1)^3-2(n+1)+1$. – John Bentin Jul 20 '15 at 17:50
Oh, thanks for catching that typo. I should have written $\dfrac{n^3+3n^2+2n}{6}$ instead of $\dfrac{n^3+3n^2+n}{6}$. I'll fix that. – JimmyK4542 Jul 20 '15 at 17:52
And yes, I agree that there is not a unique way to assign a finite value to a divergent sum, which is what the OP was asking for. – JimmyK4542 Jul 20 '15 at 17:59
Following your corrected version, $n^3+3n^2+2n=(n+1)^3-(n+1)$, which leads to the end result $\frac{11}{720}$. – John Bentin Jul 20 '15 at 18:08
@Jimmy If only the correct techniques are used, all of them will agree. I assume the OP really asked for such a value. – Vincenzo Oliva Jul 20 '15 at 18:10
"Correct" means of course "suitable for the series at issue", in my last comment. – Vincenzo Oliva Jul 20 '15 at 18:26
Don't worry, you can add, subtract and shift "infinite" series. Those interested should read http://arxiv.org/pdf/1209.5739.pdf – orion Jul 21 '15 at 12:10
@orion No, thanks. If that were true, then $$ -\frac{1}{12} "=" \sum_{n=1}^\infty n = \sum_{n=0}^\infty (n+1) = \sum_{n=0}^\infty n +\sum_{n=0}^\infty 1 "=" 0 - \frac{1}{12} -\frac{1}{2}.$$ – Vincenzo Oliva Jul 22 '15 at 08:26

Lucian · Answer 2 · 2015-07-21T11:36:30.177

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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.

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.

edited Jul 21 '15 at 11:36

answered Jul 20 '15 at 17:21

Lucian

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The first "hint" happily yields the sum $+\infty$. – Did Jul 20 '15 at 17:35
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Well, @Did, in case you haven't already noticed, all those other nice people who posted an answer on this thread, pointing to possible finite values for this divergent sum got downvoted, so what's the problem now ? Divergent values for divergent sums are also forbidden ? Can anyone win at this game ? – Lucian Jul 20 '15 at 17:45
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Not sure your comment is relevant (and you can store your guns). The OP asks for a way to attribute a finite value to the infinite sum we know and your hint explains a way to attribute it the value +oo, is that it? Some remarks: 1. This is not what the OP is asking for. 2. No need to do what you suggest to attribute the value +oo to this series. 3. The way you wrote your answer is frankly misleading, pointing clearly at a finite value. – Did Jul 20 '15 at 17:54
@Did: My last comment, though formally addressed to you, wasn't actually aimed at you... And yes, I do know what I have to do, but the fact that the other two answers that dared mention the $\zeta$ function got downvoted and deleted does not exactly inspire much in the way of encouragement. – Lucian Jul 20 '15 at 20:21
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You know what? My comments addressed to you are... well, addressed to you (so much simpler than your dances around) and they do make some specific mathematical points. That you are completely avoiding to address any of them is telling. – Did Jul 20 '15 at 21:01
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@Did: I know that I have to address $\zeta$-regularization in terms of the Dirichlet $\eta$ function. Which I won't do, because there's no point. I'd rather delete the post, than write novels, only to have them downvoted anyway, for no reason other than human cretinism, and the OP will probably delete the question anyway, also because of the downvotes. – Lucian Jul 20 '15 at 22:31
All notion of cretinism aside, is your post answering the question or not? I believe not (and I find tiring your evasion tactics trying to prevent us to examine this veeeery simple point). – Did Jul 20 '15 at 22:50
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@Did: Mistakes are easily correctable. But sometimes there is absolutely no point in doing that. All posts that ask for regularized values get downvoted for no reason. Even when both asker and answerer make it perfectly clear what they are talking about, and in precisely what way their words are to be understood. This is a recurrent and inexcusable behaviour. – Lucian Jul 20 '15 at 23:09
You mean, your post is a mistake? I agree--but then, you might want to mention the fact clearly. (The rest of your comment is fantasy, unrelated to the mathematics at hand, hence uninteresting.) – Did Jul 21 '15 at 07:09
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@Did: No, I'm afraid my comment is pretty spot-on. – Lucian Jul 21 '15 at 10:45

score 0 · Answer 3 · answered Jul 20 '15 at 17:15

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 $.

score -1 · Answer 4 · answered Jul 20 '15 at 17:05

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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 $-1=\infty$.

answered Jul 20 '15 at 17:05

user2520938

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If what you mean is that weak souls are fascinated by this kind of formulas for bad reasons, I fully agree. But there are some mathematics behind these formulas, see https://en.wikipedia.org/wiki/Ramanujan_summation. – Did Jul 20 '15 at 17:07
@Did Hm thanks for pointing that out. I've never seen it 'proven' using such techniques though; I think most people have been shown the zeta-function reasoning. Still my point is that this result is very subtle and that you have to realise that it is not the result of any 'simple'/'ordinary' manipulation of series (at least not of correct manipulations, I've also seen some 'proves' by people who've never heard of absolute convergence). – user2520938 Jul 20 '15 at 17:22
As a generalization, you are absolutely free to assign a value to infinite sum, as long as some axioms for summation hold. Check out http://arxiv.org/pdf/1209.5739.pdf – orion Jul 21 '15 at 12:11

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

4 Answers4