Why is this trollmath wrong?

Hi forum,

I saw this trollmath on youtube. i dont claim its correct, but i cant understand where the mistake is. help :)

Consider a geometric series as follows:

$\displaystyle S = 2 + 4+ 8 + 16 + ......$ [1]

$\displaystyle 2S = 4 + 8+ 16 + ...... $ [2]

[2]-[1]

$\displaystyle S = 2 $

Clearly the last line is false, but where is the mistake?

Re: Why is this trollmath wrong?

Quote:

Originally Posted by

**SpringFan25** I saw this trollmath on youtube. i dont claim its correct, but i cant understand where the mistake is. help Consider a geometric series as follows:

$\displaystyle S = 2 + 4+ 8 + 16 + ......$ [1]

$\displaystyle 2S = 4 + 8+ 16 + ...... $ [2]

[2]-[1]

$\displaystyle S = 2 $ Clearly the last line is false, but where is the mistake?

You are only allowed to subtract finite sums. SO

$\displaystyle S_n = 2 + 4+ 8 + 16 + ...+2^n$ [1]

$\displaystyle 2S_n = 4 + 8+ 16 + ....+2^{n+1}$ [2]

$\displaystyle 2S_n-S_n=2^{n+1}-2$

Re: Why is this trollmath wrong?

That is not exactly correct. You **can** subtract infinite sums if they converge. For example, 1+ 1/2+ 1/4+ 1/8+ ... (adding 1 over powers of 2) is a "geometric series" that converges to 1/(1- 1/2)= 2. Similarly, 1+ 1/3+ 1/9+ 1/27+ ... (adding 1 over powers of 3) is a geometric series that converges to 1/(1- 1/3)= 3/2. (1+ 1/2+ 1/4+ 1/9+...)- (1+ 1/3+ 1/9+ 1/27+ ...)= (1/2- 1/3)+ (1/4- 1/9)+ (1/8- 1/27)+ ...= 1/6+ 5/36+ 19/216+ ... converges to 2- 3/2= 1/2.

The problem here is that the series does not **converge**.

Re: Why is this trollmath wrong?

Quote:

Originally Posted by

**HallsofIvy** That is not exactly correct. You **can** subtract infinite sums if they converge.

Of course that is true. But the point is that to prove convergence one looks at finite sums.

Re: Why is this trollmath wrong?

Hello, SpringFan25!

Quote:

Consider a geometric series as follows:

. . $\displaystyle \begin{array}{ccccc}S &=& 2 + 4 + 8 + 16 + \hdots & {\color{red}[1]} \\ 2S &=& \quad\;\; 4 + 8 + 16 + \hdots & {\color{red}[2]} \end{array}$

$\displaystyle {\color{red}[2 ] - [1]}:\;S = -2$

Clearly the last line is false, but where is the mistake?

The mistake is in thinking that we can "do" arithmetic with Infinity.

Clearly: .$\displaystyle S \:=\:2 + 4 + 8 + 16 + \hdots \:=\:\infty$

Then: .$\displaystyle 2S \:=\:\infty$

Hence: .$\displaystyle 2S - S \:=\:\infty-\infty$, an indeterminate form.

That is, $\displaystyle S$ could be *any* value.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here is a classic "paradox".

$\displaystyle \text{Evaluate: }\:S \;=\;1 - 1 + 1 - 1 + 1 - 1 + \hdots$

Method 1

$\displaystyle S \;=\;(1-1) + (1-1) + (1-1) + \hdots$

$\displaystyle S \;=\;0 + 0 + 0 + \hdots$

$\displaystyle \boxed{S \;=\;0}$

Method 2

$\displaystyle S \;=\;1 - (1-1) - (1-1) - (1-1) - \hdots$

$\displaystyle S \;=\;1 - 0 - 0 - 0 - \hdots$

$\displaystyle \boxed{S \;=\;1}$

Method 3

$\displaystyle S$ is a geometric series

. . with first term $\displaystyle a = 1$ and common ratio $\displaystyle r = \text{-}1.$

Hence: .$\displaystyle S \;=\;\frac{a}{1-r} \;=\;\frac{1}{1-(\text{-}1)}$

Therefore: .$\displaystyle \boxed{S \;=\;\tfrac{1}{2}}$

Re: Why is this trollmath wrong?

Quote:

Originally Posted by

**SpringFan25** Hi forum,

I saw this trollmath on youtube. i dont claim its correct, but i cant understand where the mistake is. help :)

Consider a geometric series as follows:

$\displaystyle S = 2 + 4+ 8 + 16 + ......$ [1]

$\displaystyle 2S = 4 + 8+ 16 + ...... $ [2]

[2]-[1]

$\displaystyle S = 2 $

Clearly the last line is false, but where is the mistake?

We CAN get the right answer if we are careful

$\displaystyle 2S = (2 + 2) + (4 + 4) + (8 + 8) + .... $

$\displaystyle \ \ S = \ \ \ \ \ 2 \ \ \ \ + \ \ \ \ 4 \ \ \ + \ \ \ \ 8 \ \ \ \ + ... $

Subtract vertically to get S

Just because we have the difference of two infinite divergent sums doesn't necessarily put the right answer out of reach , like Soroban said infinity minus infinity could be anything... that includes the right answer. Finding a procedure that produces the right answer may be difficult though.

:)