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Question: How will you prove that \[1 + 2 + 3 + 4 + 5 + 6 + 7 + ........\infty = \dfrac{{ - 1}}{{12}}\]?...

How will you prove that 1+2+3+4+5+6+7+........∞=−1121 + 2 + 3 + 4 + 5 + 6 + 7 + ........\infty = \dfrac{{ - 1}}{{12}}?

Explanation

Solution

We will take some assumptions and equations, and using these, we will prove that sum of natural numbers up to infinity is equal to −112\dfrac{{ - 1}}{{12}}. We will take some infinite series or some infinite patterns and we will do some mathematical operations and prove the given statement.

Complete step by step answer:
First of all, take an assumption that, A=1−1+1−1+1−1+1−1+......∞A = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ......\infty
Now, subtract this whole equation from 1.
⇒1−A=1−(1−1+1−1+1−1+1−1+......∞)\Rightarrow 1 - A = 1 - (1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + ......\infty )
⇒1−A=1−1+1−1+1−1+1−1+1−......∞\Rightarrow 1 - A = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - ......\infty
If you observe here, you get the same assumption as of A, on the right-hand side.
So, we can write,
⇒1−A=A\Rightarrow 1 - A = A
⇒1=A+A=2A\Rightarrow 1 = A + A = 2A
So, we can conclude as A=1−1+1−1+1−1+1−1+1−......∞=12A = 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1 - ......\infty = \dfrac{1}{2} ----(1)

Now, take another assumption that, B=1−2+3−4+5−6+7−8+9−......∞B = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - ......\infty
Now, subtract this equation from equation ‘A’.
⇒A−B=(1−1+1−1+1−.....∞)−(1−2+3−4+5−......∞)\Rightarrow A - B = (1 - 1 + 1 - 1 + 1 - .....\infty ) - (1 - 2 + 3 - 4 + 5 - ......\infty )
On rearranging the right-hand side by grouping corresponding terms of both series, we get another interesting pattern which is,
⇒A−B=(1−1)+(−1+2)+(1−3)+(−1+4)+(1−5)+(−1+6)+.....∞\Rightarrow A - B = (1 - 1) + ( - 1 + 2) + (1 - 3) + ( - 1 + 4) + (1 - 5) + ( - 1 + 6) + .....\infty
⇒A−B=0+1−2+3−4+5+.....∞\Rightarrow A - B = 0 + 1 - 2 + 3 - 4 + 5 + .....\infty
If you observe, the right-hand side is again equal to ‘B’.
⇒A−B=B\Rightarrow A - B = B
⇒A=2B\Rightarrow A = 2B

And we know the value ‘A’ from equation (1). So, we can write it as,
⇒12=2B\Rightarrow \dfrac{1}{2} = 2B
So, we can conclude that, B=1−2+3−4+5−6+7−8+9−......∞=14B = 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8 + 9 - ......\infty = \dfrac{1}{4} -----(2)
Now, make another assumption that C=1+2+3+4+5+6+7+8+.......∞C = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + .......\infty
Now, subtract this equation from ‘B’.
⇒B−C=(1−2+3−4+5−......∞)−(1+2+3+4+5+.......∞)\Rightarrow B - C = (1 - 2 + 3 - 4 + 5 - ......\infty ) - (1 + 2 + 3 + 4 + 5 + .......\infty )
⇒B−C=(1−2+3−4+5−......∞)−1−2−3−4−5−.......∞\Rightarrow B - C = (1 - 2 + 3 - 4 + 5 - ......\infty ) - 1 - 2 - 3 - 4 - 5 - .......\infty
On rearranging the terms, we get,
⇒B−C=(1−1)+(−2−2)+(3−3)+(−4−4)+(5−5)+........∞\Rightarrow B - C = (1 - 1) + ( - 2 - 2) + (3 - 3) + ( - 4 - 4) + (5 - 5) + ........\infty
⇒B−C=−4−8−12−16−.......∞\Rightarrow B - C = - 4 - 8 - 12 - 16 - .......\infty

Now, if we take −4 - 4 common, we get,
⇒B−C=−4(1+2+3+4+.....∞)\Rightarrow B - C = - 4(1 + 2 + 3 + 4 + .....\infty )
⇒B−C=−4C\Rightarrow B - C = - 4C
Again, on doing some transpositions, we get,
⇒B=−3C\Rightarrow B = - 3C
And we know the value of ‘B’ from equation (2). So, on substitution, we get,
⇒14=−3C\Rightarrow \dfrac{1}{4} = - 3C
⇒C=−112\Rightarrow C = \dfrac{{ - 1}}{{12}}
So, we can conclude that, C=1+2+3+4+5+6+7+......∞=−112C = 1 + 2 + 3 + 4 + 5 + 6 + 7 + ......\infty = \dfrac{{ - 1}}{{12}}
And in this way, we can prove this statement.

Note: Take care while grouping the terms of different series and also be careful with the signs, because even one single sign can change the whole equation and your proof may go wrong. All the assumptions that we took here are infinite assumptions or infinite patterns and make a note that ∞−∞≠0\infty - \infty \ne 0. Actually, the phrase ∞−∞\infty - \infty is not defined.