Infinity Is Negative?

Calculus Level 1

I'm going to prove that 1 + 2 + 2 2 + 2 3 + ⋯ = − 1 1 + 2 + 2^2 + 2^3 + \cdots = -1 . Is the following working correct?

Let A A denote the value of 1 + 2 + 2 2 + 2 3 + ⋯ 1 + 2 + 2^2 + 2^3 + \cdots , then

2 A = 2 + 2 2 + 2 3 + 2 4 + ⋯ 2 A + 1 = 1 + 2 + 2 2 + 2 3 + 2 4 + ⋯ 2 A + 1 = A A = − 1. \begin{aligned} 2A &=& 2 + 2^2 + 2^3 + 2^4 + \cdots \\ 2A + 1 &=&1 + 2 + 2^2 + 2^3 + 2^4 + \cdots \\ 2A + 1 &=&A \\ A &=&-1. \end{aligned}

Yes, it is correct No, it is not correct

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Pi Han Goh by Pi Han Goh , 30, Malaysia

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5 solutions

I want to say that this is not a formal solution. Having said this, I guess what is wrong is that because the series doesn't converges it cannot be say " A A denote the value of 1 + 2 + 2 2 + 2 3 + ⋯ 1 + 2 + 2^2 + 2^3 + \cdots " cause there's no such value

47 Helpful 4 Interesting 4 Brilliant 1 Confused

Moderator note:

Note that the series 1 + 2 + 2 2 + 2 3 + ⋯ 1+2+2^2+2^3+ \cdots diverges so we can't subtract both sides by infinity.

I think the ans makes some sense. it is similar to the sum 1+2+3+4+...=-1/12. This famous -1/12 is actually found in physics calculations and its uses the Riemann zeta function to find sums to infinity by eliminating the infinity part and keeping the finite part. Correct me.

Alex Loy - 4 years, 8 months ago

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wrong. we can subtract by ∞.

Am Kemplin - 2 weeks, 5 days ago

To say that the infinite sum "converges" for Re(s) > 1 is to say that the complex numbers represented by finite partial sums of the form

SN(s) = ∑1≤n<N 1/ns have a well-defined limiting value, for each s with Re(s) > 1, as N becomes arbitrarily large.

Alex Loy - 4 years, 8 months ago

But, it's possible according to Euler's work on diverging series.

Ahmed Imtiaz - 3 years, 3 months ago

If this explanation is wrong then even the ramanujan explanation for sum of natural nos. =-12 is also wrong???

Ayush Singh - 1 year, 10 months ago

Thien-Lee@blue-dragons-co.org had visit Pluto planet life form misconceptions.

Thien Canh Lee - 1 year ago
Rishik Jain
Aug 12, 2016

Since, the infinite series is a Geometric Progression with r > 1 r \gt 1 , the series diverges. And we can't subtract by a series that does not converge to a finite sum.

22 Helpful 2 Interesting 2 Brilliant 0 Confused
Loillipop Cuber
Mar 19, 2018

Cannot manipulate diverging infinite series. Other wise the sum of all natural numbers is -1/12...;)

6 Helpful 1 Interesting 0 Brilliant 0 Confused
Hosein Jahanshahi
Aug 11, 2016

Well I think it's not logical that a series of sums will equal negative. Even we use this fact in other solutions.

3 Helpful 0 Interesting 0 Brilliant 1 Confused

Lol ik right

Jerry Christensen - 2 years, 7 months ago
Kőszeghy Csaba
Jan 25, 2019

We know that SUMn(1/q^n)=1/(1-q) if |q|<1. Plugging in 2 we get -1.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

But we can't plug in 2 because |q| < 1 is not fulfilled, right?

Pi Han Goh - 2 years, 4 months ago

x² × 68 = 2z

Am Kemplin - 3 weeks ago

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