Is it true??\textbf{Is it true??}n=1n\displaystyle\sum_{n=1}^{\infty}n\neq\infty

In a series of summation if each number is bigger than the previous one then the sequence is called as diverging series. For example :n=12n=2+4+6+8+10+12+\displaystyle\sum_{n=1}^{\infty}2n=2+4+6+8+10+12+\ldots this is a diverging serise.

But does all the diverging series sum up to infinity\textbf{infinity} ????? For example S=1+2+3+4+5+6+7+8+S=1+2+3+4+5+6+7+8+\ldots is a diverging series does not sum up to Infinity.

Here is the proof :

Let,

S=1+2+3+4+5+6+7+8+S=1+2+3+4+5+6+7+8+\ldots equation1\text{equation}\boxed{1}

S1=11+11+11+11+S_1=1-1+1-1+1-1+1-1+\ldots equation2\text{equation}\boxed{2}

S2=12+34+56+78+S_2=1-2+3-4+5-6+7-8+\ldots equation3\text{equation}\boxed{3}

Now Subtracting equtation 2\boxed{2} from 1\boxed{1}.

1S1=1(11+11+11+1)1-S_1=1-(1-1+1-1+1-1+1-\ldots)

1S1=11+11+11+11+1-S_1=1-1+1-1+1-1+1-1+\ldots

1S1=S11-S_1=S_1

1=S1+S11=S_1+S_1

2S1=12S_1=1

S1=12\boxed{S_1=\frac{1}{2}}

Adding equation 3\boxed{3} to equation 3\boxed{3}.

S2+S2=(12+34+56+78+)+(12+34+56+78+)S_2+S_2=(1-2+3-4+5-6+7-8+\ldots)+(1-2+3-4+5-6+7-8+\ldots)

2S2=1+(2+1)+(32)+(4+3)+(54)(6+5)+(78)2S_2=1+(-2+1)+(3-2)+(-4+3)+(5-4)-(6+5)+(7-8)-\ldots

2S2=11+11+11+12S_2=1-1+1-1+1-1+1-\ldots

2S2=S12S_2=S_1

2S2=122S_2=\frac{1}{2}

S2=14\boxed{S_2=\frac{1}{4}}

Subtracting equation 1\boxed{1} from equation 3\boxed{3}.

S2S=(12+34+56+78+)(1+2+3+4+5+6+7+8+)S_2-S=(1-2+3-4+5-6+7-8+\ldots)-(1+2+3+4+5+6+7+8+\ldots)

S2S=(11)+(22)+(33)+(44)+(55)+(66)+(77)+(88)+S_2-S=(1-1)+(-2-2)+(3-3)+(-4-4)+(5-5)+(-6-6)+(7-7)+(-8-8)+\ldots

S2S=04+08+012+016+020+024+028+032+S_2-S=0-4+0-8+0-12+0-16+0-20+0-24+0-28+0-32+\ldots

S2S=48121620242832S_2-S=-4-8-12-16-20-24-28-32-\ldots

S2S=4(1+2+3+4+5+6+7+8+)S_2-S=-4(1+2+3+4+5+6+7+8+\ldots)

14S=4S\frac{1}{4}-S=-4S

14=4S+S\frac{1}{4}=-4S+S

14=3S\frac{1}{4}=-3S

S=1(4)(3)S=\frac{1}{(4)(-3)}

S=112S=\frac{1}{-12}

S=112\boxed{S=\frac{-1}{12}}

1+2+3+4+5+6+7+8+=112\boxed{1+2+3+4+5+6+7+8+\ldots=\frac{-1}{12}}

Thus sum of all natural numbers is 112\frac{-1}{12}

I just want to know is this proof mathematicaly correct. Please reply and share your views.You can aslo state any of the intresting proofs or properties or anything related to maths that you find intresting below..\downarrow

Note by Sid 2108
4 years, 8 months ago

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I think the way you have shown it is interested, but a classic fallacious proof from the principle below: +=2 \infty + \infty = \infty \ne 2 \infty In other words, you can't sum and subtract divergent series because the value of the overall expression will depend on the order of addition and subtraction. However, the Riemann Zeta Function has been used to show that the sum of all natural numbers is indeed -112\frac{1}{12}, which has been used in string theory and other areas.

Curtis Clement - 4 years, 8 months ago

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A example of what I have said can be seen with the alternating harmonic series: (112+1314+15...)=(1+13+15+17+..)(12+14+16+...) (1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5}-...) = (1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7}+..) - (\frac{1}{2} + \frac{1}{4} + \frac{1}{6}+...) =(1+12+13+14+15+...)(1+12+13+14+15+...)=0= (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+...) - (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+...) = 0 which is not correct because the sum actually converges to Ln(2) by the Taylor series.

Curtis Clement - 4 years, 8 months ago

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Oh I got it. And do you know about the riemann hypothesis\textbf{riemann hypothesis} The million dollar problem which the Clay Math Institute has posted.(I just asked because you stated the zeta function)

Sid 2108 - 4 years, 8 months ago

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