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:

[1]

[2]

[2]-[1]

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

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- July 12th 2013, 06:47 AMSpringFan25Why 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:

[1]

[2]

[2]-[1]

Clearly the last line is false, but where is the mistake? - July 12th 2013, 07:01 AMPlatoRe: Why is this trollmath wrong?
- July 12th 2013, 07:46 AMHallsofIvyRe: 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**. - July 12th 2013, 07:51 AMPlatoRe: Why is this trollmath wrong?
- July 12th 2013, 08:54 AMSorobanRe: Why is this trollmath wrong?
Hello, SpringFan25!

Quote:

Consider a geometric series as follows:

. .

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

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

Clearly: .

Then: .

Hence: . , an indeterminate form.

That is, could bevalue.*any*

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

Here is a classic "paradox".

Method 1

Method 2

Method 3

is a geometric series

. . with first term and common ratio

Hence: .

Therefore: .

- July 12th 2013, 10:54 PMagentmulderRe: Why is this trollmath wrong?
We CAN get the right answer if we are careful

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.

:)