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# Infinite Sums and the Peculiarities of Mathematics

Written by Raymond Santos Estrella on Saturday, 18 January 2014. Posted in 2014

Back in high school when I was at the peak of my love for mathematics, I was enamored with the concept of infinite series. Related to this are factorials, arithmetic sequences, and geometric sequences, and the subject of this post: the infinite sum.

One of the more interesting things I got to discover was the proof that the infinite sum of all rational positive integers was not $∞$, but rather $-112$. Yeah, it’s quite true and I’ll prove it shortly. You may not believe me at first as it seems inconceivable that adding all possible positive integers would result in such a small number, much less a negative one. However, bear with me for a few minutes and I’ll show you the same proof that I submitted for a paper in algebra back in my senior year of high school. Believe me, I spent a few days checking, leaving it alone, then coming back and rechecking my calculations again and I came up with the same answer every time. I thought I had come up with something utterly new and unique but unfortunately, I was beaten to it by a couple decades. Srinivasa Ramanujan, of whom I’ve written about before, also came up with the same intuitive proof in 1913. He wrote to G. H. Hardy:

Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich’s Infinite Series and not fall into the pitfalls of divergent series. … I told him that the sum of an infinite number of terms of the series: $1+2+3+4+…=-112$ under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter. …

First, let me restate the problem above. I’ll write it as the series $S$:

$S=1+2+3+4+5+6+7+8+9+\dots$

In sigma notation we would write it this way:

$\underset{n=1}{\overset{\infty }{\sum k}}\frac{n\left(n+1\right)}{2}$

And for purposes of this post, I am going to prove the validity of this series:

$\sum _{n=1}^{\infty }k\to -\frac{1}{12}$

Cool, right? There are six parts to this proof and I’ll go through them step by step and try to be as verbose as I can. Don’t worry, you don’t need an advanced degree in mathematics, just the most basic arithmetic and algebra skills are needed. However, if you don’t get a certain part of the math, feel free to comment below or drop me a line and I’ll get back to you as soon as I can.

### Definining the series

Actually, before all this, let’s get some technical terms out of the way. When numbers are added one after another to find a sum, this is called a summation and expresed algebraically with the summation symbol which uses the capital Greek letter sigma (Σ). You can add upper and lower limits above and below the Σ symbol, respectively, and then write the pattern for creating the sum afterward. This works well for a finite set of values. On the other hand, when we try to find the sum of an infinite sequence of numbers (called an infinite series), actual computation becomes practically impossible. Sometimes, depending on the series, we can come up with a particular value (i.e. $0+0+0+0+…=0$), no value at all, or a convergent value as I’ll be showing you below.

To start us off, I’ll define the three different arithmetic series that we’ll be using for this proof. Why I chose to use these three is a bit arbitrary on my part. When I had to do this on paper so long ago, I had to work backwards then forwards in a sort of trial and error system that can only loosely be called a “method”. In any case, I used the series $S$, $S1$, and $S2$ defined as follows:

$S=1+2+3+4+5+6+7+8+9+\dots$

and

${S}_{1}=1-1+1-1+1-1+1-1+1-\dots$

and

${S}_{2}=1-2+3-4+5-6+7-8+9-\dots$

### Evaluating $S1$

First, we’ll evaluate $S1$. This is actually a pretty complex problem which—spoiler alert—will actually give us three very different answers.

Note: Is it really possible for a single equation to net more than one answer? Yes! When dealing with infinite arithmetic series, it is possible to have more than one value in a solution set. The existence of one value does not make any other value incorrect. Relax! haha

$S1$, as you’ll remember is written this way:

${S}_{1}=1-1+1-1+1-1+1-1+1-\dots$

or, in sigma notation:

$∑n=0∞-1n$

If we were to take a calculator and monitor the running sum, the answer would oscillate between 1 and 0, depending on whether we stopped at an odd or even step. So which one is the answer? We can’t really say for sure since we’re supposed to be performing the computation ad infinitum. While it seems that we’re stuck since we can’t feasibly do the actual calculation up to infinity and until the end of time, we can facilitate computation by inserting parenthesis and grouping terms together. Since addition is both associative and commutative, the insertion would not affect the end sum. Let’s try this pattern first:

${S}_{1}=\left(1-1\right)+\left(1-1\right)+\left(1-1\right)+\left(1-1\right)+\dots$

Performing the arithmetic within each parenthesis will result in 0. Rewriting it gives us this new series:

${S}_{1}=0+0+0+0+\dots$

Now we know that $S1=0$.

However, we’re not quite finished. If we shift the brackets to the right one step and group the terms differently, and owing to the associative and commutative properties of addition, we should come up with the exact same value, right? Let’s see:

${S}_{1}=1+\left(-1+1\right)+\left(-1+1\right)+\left(-1+1\right)\dots$

Performing the arithmetic in parenthesis gives results in 0. However, note the initial 1 that wasn’t included in the parenthesis:

${S}_{1}=1+0+0+0+\dots$

Thus we have the strange situation where $S1$ can be expressed as both $S1=0$ and $S1=1$, depending on how you compute the series.

### Introducing Grandi’s series

Now, how do we reconcile this seeming paradox? Well, the Italian mathematician, philosopher and priest Guido Grandi found a novel solution back in the early 18th century. While it was initially met with skepticism and a lot of head scratching, his idea became mainstream after a couple hundred years. Today, it’s all but accepted that his solution is probably the correct one.

Grandi started off by making a new equation: $1-S1$. Thus:

$1-S1=1-1-1+1-1+1-1+1-1+1-…$

Reversing all signs inside the parenthesis gives us:

$1-S1=1-1+1-1+1-1+1-1+1-…$

See anything familiar? Everything at the right side of the equation is identical to the original statement of ${S}_{1}$. If we were to substitute the ${S}_{1}$ variable for that string, we get this equation:

${1-S}_{1}={S}_{1}$

Let’s group the similar terms by keeping the 1 on the left and moving the ${S}_{1}$ to the other side.

$1={S}_{1}+{S}_{1}$ $1=2{S}_{1}$

Divide both sides by 2 and we finally get the third answer: $\frac{1}{2}$.

${S}_{1}=\frac{1}{2}$

Now that we have three different answers for $S1$, aren’t we triply perplexed? Not really. You see, both 1 and 0 are quite valid possible answers for this infinite series. If we didn’t employ parenthesis, we wouldn’t be able to come up with an answer at all since we’d have to calculate ad infinitum which is practically impossible. Thus, the employment of parenthesis was necessary in order to come up with a possible answer. But how about $12$?

Historical thought on this problem has shown that the great body of mathematicians have come to accept that $12$ is the most likely answer to this algorithm. This is because of the great difficulty in dealing with infinite series, mathematicians have had to make do with alternative ways of solving problems such as this. If we were to graph this as we go step by step in calculating $S1$ the traditional way, you’ll see that the answer converges between 0 and 1, specifically at $12$. It acts the same way that a limit does but the curve is not asymptotic in that it will get infinitely closer to $12$ but never cross it. Rather, it actually will converge at $12$ an infinite number of times as it oscillates between the original answers 0 and 1. Grandi recognized this immediately and further published his finding that the answer is most likely $12$.

Note that for purposes of the rest of the calculations that follow, we will assume that Grandi is correct and that $S1$ will give a value of $12$.

### Calculating $S2$

Moving along, we’ll now solve for $S2$. Again, it may seem arbitrary but to facilitate computation, we’ll be doubling $S2$. Furthermore, I’ll write it differently by shifting the addend to the right to show a strange phenomena when combining the numbers. No hocus-pocus here. Remember that multiplication is just addition at it’s most basic level and that in addition, associative and commutative properties apply and thus, the order in which we perform the calculations or the position of the values are irrelevant to the answer. We’re going to get the same sum anyway, this is just to make calculation and grasping the infinite series much easier to perform.

$\begin{array}{ccccccccccccc}2{S}_{2}& =& 1& -2& +3& -4& +5& -6& +7& -8& +9& -\dots & \\ & +& & +1& -2& +3& -4& +5& -6& +7& -8& +\dots & \\ & =& 1& -1& +1& -1& +1& -1& +1& -1& +1& -\dots & \end{array}$

Notice here that the sum results in exactly the same string as $S1$. In that case, we can restate the equation into:

$2{S}_{2}={S}_{1}$

If we use Grandi’s answer for $S1$ and plug in $12$ , we get:

$2{S}_{2}=\frac{1}{2}$

Now we divide both sides by 2 and are left with:

${S}_{2}=\frac{1}{4}$

### Evaluating $S-{S}_{2}$

Now we’re getting somewhere aren’t we! Again, seemingly arbitrary choice of equations, but bear with me a little bit more. You’ve made this far anyway. Let’s evaluate this:

$S-{S}_{2}$

In long subtraction form, we get something like this:

$\begin{array}{cccccccccccccccccccccc}S& =& 1& +& 2& +& 3& +& 4& +& 5& +& 6& +& 7& +& 8& +& 9& +& \dots & \\ {S}_{2}& =-\left[& 1& -& 2& +& 3& -& 4& +& 5& -& 6& +& 7& -& 8& +& 9& -& \dots & \right]\\ S-{S}_{2}& =& 0& +& 4& +& 0& +& 8& +& 0& +& 12& +& 0& +& 16& +& 0& +& \dots & \end{array}$

Now notice that the sum is 0 on odd steps of the series while there is a non-zero sum on the even steps. Furthermore, take note that the non-zero sums are divisible by four. We can thus simplify this pattern this way:

$S-{S}_{2}=4\left(1+2+3+4+\dots \right)$

Now look at that string $1+2+3+4+…$. Isn’t it familiar? Yeah, you guessed it! It’s actually $S$! We can now restate this equation like this:

$S-{S}_{2}=4S$

### Whew, this was tiring

Now, before we proceed with the final step my brilliant proof, let’s recap the values that we’ve found earlier and write them down here.

$S1=12$

and

$S2=14$

Going back to our math and replacing $S2$ with $14$ gives us:

$S-\frac{1}{4}=4S$

Grouping similar terms, we get:

$S-4S=\frac{1}{4}$ $-3S=\frac{1}{4}$

Dividing both sides by -3 ultimately nets us:

$S=-\frac{1}{12}$

### Raymond Santos Estrella

I guess I should really make a proper writeup here. Something witty or maybe a joke to add some levity. I’ll come back to this when I have time. If you have any suggested copy that I can insert here, drop me a line.

• 18 January 2014 at 07:38 |

Mind. Blown.

• 18 January 2014 at 08:00 |

i'm flipping out from how awesome this is!!

• 18 January 2014 at 08:57 |

is dis for real?? how can it possibly be -1/12???

• 18 January 2014 at 09:18 |

Brilliant post Ray! But really, how did you even find this out in the first place? What high school kid is that curious about math?! Lol

• 18 January 2014 at 09:28 |

I know! I was weird in that way! I liked basketball and Michael Jordan and watched MTV so I guess it all balances out.

• 18 January 2014 at 09:38 |

Wow! I have to admit that seeing all that math kind of intimated me but you explained it pretty well and so simply. It still boggles my mind how the sum turned out that way, though.

• 18 January 2014 at 09:53 |

How did you come up with this?

• 18 January 2014 at 10:50 |

Does not compute. /end robot voice

• 18 January 2014 at 10:51 |

You found this out independently from Ramanujan? How long did that take?

• 18 January 2014 at 11:03 |

It took me about a week of playing around with the sums until I found the answer. I didn't tell anyone what I was working on at the time since I felt it was such a breakthrough lol. In hindsight, I might have save myself the effort if I looked around in the library for something like this.

• 18 January 2014 at 11:24 |

All that math! O.o

• 23 January 2014 at 16:54 |

How did you get all the equations to display that nicely?

• 23 January 2014 at 17:24 |

I'd also like to know this.

• 23 January 2014 at 18:02 |

It's a combination of MathML for the semantics and a jQuery-based Javascript plugin for the presentation. Looks really great and search engines ought to be able to understand it.

• 23 January 2014 at 18:24 |

Thanks!