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

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

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

Here is the proof :


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

\(S_1=1-1+1-1+1-1+1-1+\ldots\) \(\text{equation}\boxed{2}\)

\(S_2=1-2+3-4+5-6+7-8+\ldots\) \(\text{equation}\boxed{3}\)

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







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







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













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

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
2 years, 7 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: \[ \infty + \infty = \infty \ne 2 \infty \] 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 -\(\frac{1}{12}\), which has been used in string theory and other areas.

Curtis Clement - 2 years, 7 months ago

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

Sid 2108 - 2 years, 7 months ago

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

Curtis Clement - 2 years, 7 months ago

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