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# The sum of every positive integers

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What is the sum of every positive integers ?

1+2+3+4+5+6+...=?

EinsteinJr  May 9, 2015

#4
+869
+8

Precisely.

The answer is $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{12}}}}$$; it can seem strange, because it's a negative number, and it's not an integer; but that result is absolutely true.

There's the proof I usually use:

$$\\S_1=1-1+1-1+1-1+... \\S_2=1-2+3-4+5-6+... \\S=1+2+3+4+5+6+... \\\\S_1=1-1+1-1+1-1+... \\1-S_1=1-(1-1+1-1+1-...)=1-1+1-1+1-1+...=S_1 \\1=2S_1 \\S_1=\frac{1}{2}$$

$$\\\\S_2=1-2+3-4+5-6+... \\S_2+S_1=(1-2+3-4+5-6+...)+(1-1+1-1+1-1+...) \\S_2+S_1=2-3+4-5+6-7+... \\-1+S_1+S_2=-1+(2-3+4-5+6-7+...)=-1+2-3+4-5+6-7+...=-S_2 \\-1+\frac{1}{2}=-2S_2 \\-2S_2=-\frac{1}{2} \\S_2=\frac{-\frac{1}{2}}{-2}=\frac{1}{4}$$

$$\\\\S=1+2+3+4+5+6+... \\S-S_2=(1+2+3+4+5+6+...)-(1-2+3-4+5-6+...)=4+8+12+16+20+24... \\=4(1+2+3+4+5+6+...)=4S \\-S_2=3S \\-\frac{1}{4}=3S \\S=-\frac{-\frac{1}{4}}{3}=-\frac{1}{12}$$

EinsteinJr  May 9, 2015
#1
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Well....we couldn't calculate the sum of all of them....!!!  {the sum would be infinite !!!}

But...we have a "formula" to calculate the sum of the first N of them.....

Let's take a simple example to see if we can find out what it is.......

Notice that.......if we had an some number of integers  to add....we might have something like this.....

1 + 2 + 3 + .....+ N-2 + N-1 + N

Now, let's rewrite the same sum backwards and add both things......so we have

[ 1              + 2             +     3 +.......         + N-2         + N-1           + N ]

+ [N              + N-1         +    N-2 +    .....      + 3             +  2             + 1]

= [N + 1]     + [N + 1]    + [N + 1] +  .....      +[N + 1]    +[N + 1]     +[ N + 1]

Now, notice that we have "N" pairs of  [N + 1] terms.....so the sum of these is just  ...  N(N+ 1)

But.......we've summed the series twice.....so we need to divide the above result by 2

And we arrive at the "formula" for the sum of the first N integers......N(N + 1) / 2    .......  !!!!

CPhill  May 9, 2015
#2
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OK, but this works only for a finite amount of numbers; now, there are infinitely many numbers from 1 to infinity.

But the correct answer is not infinity; it's a real number. And it's just astounding, I think.

EinsteinJr  May 9, 2015
#3
+8

The answer is actually $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{12}}}}$$

Let:

S=1+2+3+4+5+6+7+8+9+n

We know the answer to the following infinite sum:

S2=1-1+1-1+1-1+1-1+1-1+1-1+1-....=$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$

We can therefore solve for this sum:

S3=1-2+3-4+5-6+7-8+9-....

S3=      1-2+3-4+5-6+7-8...

S3=       +1-2+3-4+5-6+7-8...

________________

2S3=    1-1+1-1+1-1+1-1+1-....

Therefore:

2S3=S2=$${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$

S3=$${\frac{{\mathtt{1}}}{{\mathtt{4}}}}$$

We can subtract S3 from S:

S-S3=  1+2+3+4+5+6+7+8+9-...

-[1- 2+3- 4+5- 6+7- 8+9-...

_________________________

4   +  8  +  12  +  16   +....

If we take a factor of 4 out:

S-S3=4(1+2+3+4+5+6+7+8+9+...)

The sum inside the parentheses is S:

S-S3=4(S)

We know S3=$${\frac{{\mathtt{1}}}{{\mathtt{4}}}}$$

S-$${\frac{{\mathtt{1}}}{{\mathtt{4}}}}$$=4(S)

Subtract S from each side:

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

Divide by 3:

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

∑ n+1=$${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{12}}}}$$

n

-Daedalus

Guest May 9, 2015
#4
+869
+8

Precisely.

The answer is $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{12}}}}$$; it can seem strange, because it's a negative number, and it's not an integer; but that result is absolutely true.

There's the proof I usually use:

$$\\S_1=1-1+1-1+1-1+... \\S_2=1-2+3-4+5-6+... \\S=1+2+3+4+5+6+... \\\\S_1=1-1+1-1+1-1+... \\1-S_1=1-(1-1+1-1+1-...)=1-1+1-1+1-1+...=S_1 \\1=2S_1 \\S_1=\frac{1}{2}$$

$$\\\\S_2=1-2+3-4+5-6+... \\S_2+S_1=(1-2+3-4+5-6+...)+(1-1+1-1+1-1+...) \\S_2+S_1=2-3+4-5+6-7+... \\-1+S_1+S_2=-1+(2-3+4-5+6-7+...)=-1+2-3+4-5+6-7+...=-S_2 \\-1+\frac{1}{2}=-2S_2 \\-2S_2=-\frac{1}{2} \\S_2=\frac{-\frac{1}{2}}{-2}=\frac{1}{4}$$

$$\\\\S=1+2+3+4+5+6+... \\S-S_2=(1+2+3+4+5+6+...)-(1-2+3-4+5-6+...)=4+8+12+16+20+24... \\=4(1+2+3+4+5+6+...)=4S \\-S_2=3S \\-\frac{1}{4}=3S \\S=-\frac{-\frac{1}{4}}{3}=-\frac{1}{12}$$

EinsteinJr  May 9, 2015
#5
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EinsteinJr and Daedalus

WHAT ARE YOU TWO DOING  ?????

The sum of every positive integer is infinity.

Melody  May 10, 2015
#6
+869
+5

You might think that all that calculation are wrong; but they AREN'T.

And if you don't trust me: https://youtu.be/w-I6XTVZXww

Note: If you still don't trust me, Hendrik Casimir used this result in his equations, and the calculation were all right.

EinsteinJr  May 10, 2015
#7
+94101
+5

Thanks EinsteinJr,

I have watched those videos and they are interesting but they start out with premises that they do not even attempt to prove.  Not on those clips anyway.

Why do you accept that 1-1+1-1+....... = 0.5    ?      (Edited)

Melody  May 10, 2015
#8
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The 1st sum is 1-1+1-1+1-1+... and not 1+0+1+0+1+0+...

This sum is called the Grandi's Series, and it's equal to $${\frac{{\mathtt{1}}}{{\mathtt{2}}}}$$ . Watch this video for more information: https://www.youtube.com/watch?v=PCu_BNNI5x4

EinsteinJr  May 10, 2015
#9
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Yes i watched the video and it was very entertaining

Melody  May 10, 2015
#10
+869
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That's why I love Numberphile.

EinsteinJr  May 10, 2015