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(The Solution Is 1)
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Below is how Grandi found 1:
 
Below is how Grandi found 1:
   
He wrote out the sum in question... hehe yay
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He wrote out the sum in question...
 
:<math>1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1...</math>
 
:<math>1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1...</math>
   
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It seems we've hit a roadblock; the method of finding partial sums doesn't work with Grandi's series. How can we show why most mathematicians prefer the answer to be ½? Thankfully, there's another way to find solutions to infinite sums.
 
It seems we've hit a roadblock; the method of finding partial sums doesn't work with Grandi's series. How can we show why most mathematicians prefer the answer to be ½? Thankfully, there's another way to find solutions to infinite sums.
   
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Let's go back to
Let's go back to <math>\sum_{x=0}^{\infty\,}\frac{1}{2^x}\,</math> and its partial sums, which are 1, 1.5, 1.75, 1.875, etc. If we take the [[mean]] of these partial sums, we will get a series that also tends towards 2. The first of these averages is 1, so let's find the others.
 
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<math>\sum_{x=0}^{\infty\,}\frac{1}{2^x}\,</math>
 
and its partial sums, which are 1, 1.5, 1.75, 1.875, etc. If we take the [[mean]] of these partial sums, we will get a series that also tends towards 2. The first of these averages is 1, so let's find the others.
   
 
:<math>\frac{1}{2}\,(1 + \frac{3}{2}\,) = \frac{1}{2}\, + \frac{3}{4}\, = \frac{2}{4}\, + \frac{3}{4}\, = \frac{5}{4}\, = 1.25</math>
 
:<math>\frac{1}{2}\,(1 + \frac{3}{2}\,) = \frac{1}{2}\, + \frac{3}{4}\, = \frac{2}{4}\, + \frac{3}{4}\, = \frac{5}{4}\, = 1.25</math>

Latest revision as of 22:06, November 29, 2019

Grandi's series is an infinite sum in mathematics. The sum was found to be divergent, or without a "true" value, in 1703 by Italian mathematician, philosopher, and priest Guido Grandi.

The SeriesEdit

Grandi's series can be written as an infinite string of alternating positive and negative 1's:

$ 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1... $

Grandi's series can also be written simply using the summation Σ:

$ \sum_{x=0}^{\infty\,}(-1)^x $

The SolutionsEdit

Grandi found three solutions to the infinite sum: 0, 1, and a mystery answer that will be revealed later.

The Solution Is 0Edit

Below is how Grandi found 0:

He wrote out the sum in question...

$ 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1... $

...he placed brackets around each 1 - 1...

$ (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1)... $

...solved each bracket...

$ 0 + 0 + 0 + 0 + 0 + 0... $

...and found the sum to be 0.

$ 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1... = 0 $

The Solution Is 1Edit

Below is how Grandi found 1:

He wrote out the sum in question...

$ 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1... $

...he placed brackets around each -1 + 1 and changed the signs as needed...

$ 1 + (- 1 + 1) + (- 1 + 1) + (- 1 + 1) + (- 1 + 1) + (- 1 + 1) + ... $

...solved each bracket...

$ 1 + 0 + 0 + 0 + 0 + 0... $

...and found the sum to be 1.

$ 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1... = 1 $

The Solution Is...Edit

Below is how Grandi found the mystery solution:

He wrote out the sum in question and set it equal to S...

$ S = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1... $

...subtracted 1 from S...

$ 1 - S = 1 - (1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1...) $

...changed the signs as needed...

$ 1 - S = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1... $

...and found S from this operation, so, therefore...

$ 1 - S = S $

Using basic algebra,...

$ 1 - S = S $
$ 1 = 2S $
$ \frac{1}{2}\, = S $

He got ½ as the third solution. The mystery solution is ½. Mathematicians were initially as skeptical about this as you might be right now, but, when they saw Grandi's proofs, as well as what I'm about to show you, they were actually quite convinced that this could be a possible "solution" to this series.

Which Solution Is Correct?Edit

Grandi's series has been a subject of debate since 1703, and it still very much is to this day. However, the majority of mathematicians agree that the solution should be ½ rather than 0 or 1. Why is this so? Is there some legitimate reason for this, or are they just plain barmy (like they usually are)? Well, let's delve into this reasoning a little bit before we can say which is the case.

Let's talk about partial sums. These are the sums of a number of consecutive terms in a sum, infinite or finite regardless.

Allow me to show you another infinite sum to demonstrate partial sums.

$ 1 + \frac{1}{2}\, + \frac{1}{4}\, + \frac{1}{8}\, + \frac{1}{16}\, + \frac{1}{32}\, + \frac{1}{64}\, + \frac{1}{128}\, + \frac{1}{256}\, + ... = 2 $

or

$ \sum_{x=0}^{\infty\,}\frac{1}{2^x}\, = 2 $

As shown above, the solution of the infinite sum is 2. However, you can get the same answer by finding its partial sums.

Let's find the first four partial sums. The first partial sum is 1, so let's find the others.

$ 1 + \frac{1}{2}\, = \frac{2}{2}\, + \frac{1}{2}\, = \frac{3}{2}\, = 1.5 $
$ \frac{3}{2}\, + \frac{1}{4}\, = \frac{6}{4}\, + \frac{1}{4}\, = \frac{7}{4}\, = 1.75 $
$ \frac{7}{4}\, + \frac{1}{8}\, = \frac{14}{8}\, + \frac{1}{8}\, = \frac{15}{8}\, = 1.875 $

In order, these sums are: 1, 1.5, 1.75, and 1.875.

As you can see, the partial sums are tending towards 2, though they don't quite reach 2 until x is infinity. This is known as a convergent sum because the values reach a certain determinable point, in this case 2. Therefore, the limit of this function is 2.

However, this method of finding partial sums doesn't work with Grandi's series.

Let's find the first four partial sums. The first partial sum is 1, so let's find the others.

$ 1 - 1 = 0 $
$ 0 + 1 = 1 $
$ 1 - 1 = 0 $

In order, these sums are 1, 0, 1, and 0.

As you can see, the partial sums do not tend towards a specific answer. Instead, they just alternate between 1 and 0 forever, never actually reaching an answer. This is why this series is a divergent sum; the values don't tend towards a single answer.

Why Do Most Mathematicians Choose It?Edit

It seems we've hit a roadblock; the method of finding partial sums doesn't work with Grandi's series. How can we show why most mathematicians prefer the answer to be ½? Thankfully, there's another way to find solutions to infinite sums.

Let's go back to

$ \sum_{x=0}^{\infty\,}\frac{1}{2^x}\, $ and its partial sums, which are 1, 1.5, 1.75, 1.875, etc. If we take the mean of these partial sums, we will get a series that also tends towards 2. The first of these averages is 1, so let's find the others.

$ \frac{1}{2}\,(1 + \frac{3}{2}\,) = \frac{1}{2}\, + \frac{3}{4}\, = \frac{2}{4}\, + \frac{3}{4}\, = \frac{5}{4}\, = 1.25 $
$ \frac{1}{3}\,(1 + \frac{3}{2}\, + \frac{7}{4}\,) = \frac{1}{3}\, + \frac{3}{6}\, + \frac{7}{12}\, = \frac{4}{12}\, + \frac{6}{12}\, + \frac{7}{12}\, = \frac{17}{12}\, = 1.41666666... $
$ \frac{1}{4}\,(1 + \frac{3}{2}\, + \frac{7}{4}\, + \frac{15}{8}\,) = \frac{1}{4}\, + \frac{3}{8}\, + \frac{7}{16} + \frac{15}{32}\, = \frac{8}{32}\, + \frac{12}{32}\, + \frac{14}{32}\, + \frac{15}{32}\, = \frac{49}{32}\, = 1.53125 $

In order, these averages are 1, 1.25, 1.41666..., and 1.53125.

As you can see, these averages also tend towards 2.

We can apply this method to Grandi's series!

As mentioned before, the partial sums of Grandi's series are an infinite alternation of 1 and 0. By finding the means of these partial sums, can we find a value being zoned in on? Let's find out. The first of these is 1, so let's find the others.

$ \frac{1 + 0}{2}\, = \frac{1}{2}\, = 0.5 $
$ \frac{1 + 0 + 1}{3}\, = \frac{2}{3}\, = 0.66666... $
$ \frac{1 + 0 + 1 + 0}{4}\, = \frac{2}{4}\, = \frac{1}{2}\, = 0.5 $
$ \frac{1 + 0 + 1 + 0 + 1}{5}\, = \frac{3}{5}\, = 0.6 $
$ \frac{1 + 0 + 1 + 0 + 1 + 0}{6}\, = \frac{3}{6}\, = \frac{1}{2}\, = 0.5 $
$ \frac{1 + 0 + 1 + 0 + 1 + 0 + 1}{7}\, = \frac{4}{7}\, = 0.571428... $
$ \frac{1 + 0 + 1 + 0 + 1 + 0 + 1 + 0}{8}\, = \frac{4}{8}\, = \frac{1}{2}\, = 0.5 $
$ \frac{1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1}{9}\, = \frac{5}{9}\, = 0.55555... $
$ \frac{1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0}{10}\, = \frac{5}{10}\, = \frac{1}{2}\, = 0.5 $

In order, these averages are 1, 0.5, 0.66666..., 0.5, 0.6, 0.5, 0.571428..., 0.5, 0.55555..., and 0.5.

As you can see, these averages tend towards ½. Even though this seems like a limit, it really is not, even though it has all of the properties of limits. It's because of the "pseudo-limit" of ½ found in Grandi's series that most mathematicians say that the solution to Grandi's series is ½.

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