You may have encountered the popular claim that \( 0.999... = 1 \), where the three dots signify that the decimal continues forever. This is a somewhat weird claim, since it would mean that mathematics is broken. There should be no way for two different numbers to have the same value. What makes it weirder is that this is quite popular claim. I've even seen mathematicians say that it's true! But is it though? One popular proof is to first denote \( S = 0.999...\) and then multiply by \(10\) to get \( 10S = 9.999...\) and subtract \( S \) from it, to get \( 10S - S = 9.000...\) and finally dividing by \(9\) yields \( S = 1.000... = 1 \) and we see that \(0.999... = 1\)! However, there's a problem. This short derivation is not strictly speaking correct. It is veeeery close to being correct, and to see why let's look at finite decimals first. Let's say that \(S = 0.999\) (note that this is not the same as \(S = 0.999...\) ). Let's do the same trick as before, so
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Sum over all natural numbers is...
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There are some really weird things in mathematics that seemingly don't make any sense. And sometimes, even if you are explained why things are the way they are, you are still left with a boggled mind.
One of those things is the deceptively simple question of what is the sum over all natural numbers, that is, the sum 1+2+3+... all the way up to infinity. One might say that this question is completely trivial, but what if I were to say that the sum is exactly -1/12?
The first reaction is usually that it doesn't make any sense, but a simple proof of this is explained in one of Numberphiles most popular videos, titled "ASTOUNDING: 1 + 2 + 3 + 4 + 5 + ... = -1/12" which you can see below
I can assure you that the Numberphile proof is completely fine, although they don't go to details. But this only raises new questions rather than answers existing ones.
We are summing positive whole numbers together, so how can the result be a negative fraction? If you take a calculator and start summing positive whole numbers together, will you end up at -1/12 after reaching infinity?
The answer is no. If it was possible to actually sum to infinity with a calculator, the result would be infinity, because this sum is a divergent series where the last term is infinite. A divergent series does not have a finite or well defined sum and there are several ways to find out whether a series diverges (first being common sense).
A very simple way to do this, is to look at which value the numbers in the sum tend towards. When summing over all positive whole numbers, the last term in the sum is infinity. If you add infinity to any finite number you get infinity.
So if the last term in the sum is infinity (and all the other terms are finite) the sum will always diverge. However, this is not necessarily true the other way around. If the last term in a summation tends towards zero, it may or may not converge. For example, the harmonic series diverges, although the last term in the sum tends towards zero.
But if the series in the Numberphile video is divergent, then the result cannot be -1/12, right? Well, it's actually not so clear cut. The thing is, the summation rules used to get that seemingly nonsensical result are not the usual rules we sum with.
Yes, the series is divergent and yes, the result is infinity if you use the usual summing rules. But it is possible to extend to a divergent series, thus bending the rules a bit.
One way is to use Cesàro summation. It is a common tool used to evaluate a divergent series and it's really quite simple. You just sum together the first n terms and divide that by n. Then you need to check if that new sum tends towards some finite constant when n goes to infinity. If it does, then that is the result of your divergent series.
Using this summation rule, you can evaluate a series that is normally divergent. For example a series where one and minus one alternate, the Grandi's divergent series.
This result was also used as a tool in the Numberphile video to arrive at the final result. This might seem a little hand wavy, but it is mathematically rigorous. You can get the same result with other methods too, for example using the Riemann zeta function.
Historically, there has been some skepticism towards this type of summing, for example Niels Henrik Abel said: "Divergent series are in general something fatal, and it is a disgrace to base any proof on them."
Call me old fashioned, but I am inclined to agree with him.
This quote is actually a little awkward nowadays, because as I have understood it, the result that the sum over all natural numbers equals -1/12 is central to string theory. Without it, you can't get all of the 26 dimensions required in bosonic string theory (I do not know about the other varieties of strings though).
Anyway, as to the question of what is the result when you sum over all natural numbers, it depends on what rules you are using. If you just punch all of the natural numbers into a calculator, you will get infinity as a result. But if you extend the conventional calculus to a divergent series and bend the rules a bit, you will get something different.
You can do things like that in mathematics, but it does not necessarily mean that it makes any physical sense.
You may have encountered the popular claim that \( 0.999... = 1 \), where the three dots signify that the decimal continues forever. This is a somewhat weird claim, since it would mean that mathematics is broken. There should be no way for two different numbers to have the same value. What makes it weirder is that this is quite popular claim. I've even seen mathematicians say that it's true! But is it though? One popular proof is to first denote \( S = 0.999...\) and then multiply by \(10\) to get \( 10S = 9.999...\) and subtract \( S \) from it, to get \( 10S - S = 9.000...\) and finally dividing by \(9\) yields \( S = 1.000... = 1 \) and we see that \(0.999... = 1\)! However, there's a problem. This short derivation is not strictly speaking correct. It is veeeery close to being correct, and to see why let's look at finite decimals first. Let's say that \(S = 0.999\) (note that this is not the same as \(S = 0.999...\) ). Let's do the same trick as before, so
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