B This CAN'T be true (Is my proof that 1=0 correct?)

[MevsEinstein]

TL;DR Summary

I used some Calculus and the Ramanujan sum to make this proof.

After learning about this formula for the sum of increasing powers - $1+p+p^2+p^3+...=1/(1-p)$ - I decided to differentiate both sides of the equation, getting: $1+2p+3p^2+4p^3+...=1/((1-p))^2$. Substituting $1$ for $p$, I get: $1+2+3+4+...=1/0$. But Ramanujan said that $1+2+3+4+...=-1/12$, so $1/0=1/-12$, meaning $0=-12$, meaning $0=1$ (dividing both sides by $-12$). There MUST be something wrong about this proof, since $0$ is NOT equal to one. May someone help me find the fallacy?

 

[anuttarasammyak]

Summary:: I used some Calculus and the Ramanujan sum to make this proof.

After learning about this formula for the sum of increasing powers - 1+p+p2+p3+...=1/(1−p) - I

This formula holds for |p|<1, neither p=1 nor larger.

 

[Drakkith]

For starters, Ramanujan did NOT say that the simple summation of the integers 1+2+3+... is equal to -1/12.
What he did was show that you could manipulate infinite series in certain ways to make sense of some mathematical formulas involving them. Without manipulating these infinite series you don't get these counterintuitive answers. After all, it is obvious to see, and trivial to prove, that summing up all the positive integers leads to an answer of infinity. So any time you see what looks like a divergent series being set equal to a finite number then you know it's either wrong or they have manipulated it in some fashion so that the answer makes sense in a certain context. Edit: Or it only makes sense for certain values of $p$, as @anuttarasammyak points out in the post above.

See this video for more info.

I can't point to the exact fallacy involving the sum of increasing powers since math is not my specialty, but I'd bet that it also involves manipulating the infinite series in a certain way. Plugging in anything greater than or equal to 1 for $p$ obviously gives you a divergent series, so there must be some sort of more complicated mathematical analysis going on here if it turns out it's equal to $1/(1-p)$.

 

[phinds]

If it some point in a mathematical process you divide by zero (even disguised as 1-p when p=1) then you can prove ANYTHING.

 

[MevsEinstein]

This formula holds for |p|<1, neither p=1 nor larger.

Substituting $-1$, you get $1-1+1-1+...=1/2$, which is a true cesaro sum.

 

[MevsEinstein]

For starters, Ramanujan did NOT say that the simple summation of the integers 1+2+3+... is equal to -1/12.
What he did was show that you could manipulate infinite series in certain ways to make sense of some mathematical formulas involving them. Without manipulating these infinite series you don't get these counterintuitive answers. After all, it is obvious to see, and trivial to prove, that summing up all the positive integers leads to an answer of infinity. So any time you see what looks like a divergent series being set equal to a finite number then you know it's either wrong or they have manipulated it in some fashion so that the answer makes sense in a certain context. Edit: Or it only makes sense for certain values of $p$, as @anuttarasammyak points out in the post above.

See this video for more info.

I can't point to the exact fallacy involving the sum of increasing powers since math is not my specialty, but I'd bet that it also involves manipulating the infinite series in a certain way. Plugging in anything greater than or equal to 1 for $p$ obviously gives you a divergent series, so there must be some sort of more complicated mathematical analysis going on here if it turns out it's equal to $1/(1-p)$.

Thanks for your answer! It's interesting that you know about Ramanujan even though Math isn't your speciality.

 

[MevsEinstein]

If it some point in a mathematical process you divide by zero (even disguised as 1-p when p=1) then you can prove ANYTHING.

I will have to find a situation where I can divide by zero first.

 

[PeroK]

I will have to find a situation where I can divide by zero first.

You already have.

 

[MevsEinstein]

You already have.

You're absolutely right.