Uncovering the Mysterious Infinite Sum | Exploring Math Forum Question

[Routaran]

I wasnt sure which math forum to ask this in so I am putting it here, apologies if its supposed to be elsewhere.

Anyway, my question has to do with an infinite sum.
Suppose we have a series:
1+2+4+8+16+32+...
each term in the series is double the previous term starting with 1

if we multiply this by the number 1 it will remain the same
1+2+4+8+16+32+... = 1(1+2+4+8+16+32+...)

also (2-1) = 1
so we can say
1+2+4+8+16+32+... = (2-1)(1+2+4+8+16+32+...)
if we expand the RHS we get
1+2+4+8+16+32+... = (2+4+8+16+32+64+...) - (1+2+4+8+16+32+...)
1+2+4+8+16+32+... = (2+4+8+16+32+64+...) + (-1-2-4-8-16-32-...)

the first part of the RHS has a set of all positive even numbers and the 2nd part of the set has a set of all negative even numbers
all the even numbers will subtract to 0 and I'm left with
1+2+4+8+16+32+... = -1

what have I done wrong? I can't figure what and where the error is.

 

[Diffy]

Hi,

Quick question for you.

Nevermind >_<

 

[coolul007]

You've treated the sum of an infinite set as a finite numberyou're saying a = infinity, b= infinity + 1, so a-b = -1, infinity +1 = infinity.

 

[tiny-tim]

Hi Routaran!

what have I done wrong? I can't figure what and where the error is.

You've subtracted two infinities from each other (because you chose a sum that doesn't converge).

Your method works fine if you choose eg 1 + 1/2 + 1/4 + …

try it! ​

 

[Routaran]

Hi Routaran!


You've subtracted two infinities from each other (because you chose a sum that doesn't converge).

Your method works fine if you choose eg 1 + 1/2 + 1/4 + …

try it! ​

thats actually very neat. both sides approach the number 2.
I guess I have a followup question then

why did multiplying by (2-1) give me the correct answer for a convergent series but not for a divergent series?
I mean I'm multiplying by 1, shouldn't the "value" remain the same?

 

[tiny-tim]

why did multiplying by (2-1) give me the correct answer for a convergent series but not for a divergent series?
I mean I'm multiplying by 1, shouldn't the "value" remain the same?

because (as coolul007 also pointed out) you're writing ∞ - ∞ !

 

[Routaran]

I still don't understand why
i mean i get what you and coolul007 are saying, ∞ - ∞ = ∞ but i don't understand how it applies to my first example.

if i was to say for example
set of all natural numbers - set of all positive even numbers
(1+2+3+4+5+...) - (2+4+6+8+...)
its obvious, even to me, that the result is still infinite because i have a set of all positive odd numbers left over.

but in the other example, the only difference is the number 1, all the other terms are identical. I don't follow exactly how to end up with an infinity

 

[Mentallic]

∞ - ∞ = ∞

Nooo that's not necessarily true. ∞ - ∞ could be anything! It could be ∞, 0, 1, -∞ or any other finite constant.
Anyway, doing arithmetic with infinities is bound for failure. There's a reason we use limits instead

if i was to say for example
set of all natural numbers - set of all positive even numbers
(1+2+3+4+5+...) - (2+4+6+8+...)
its obvious, even to me, that the result is still infinite because i have a set of all positive odd numbers left over.

Are you sure? Because with some similar manipulations as you did above, we can change the answer.

A = (1+2+3+4+5+...) - (2+4+6+8+...)
= (1+2+3+4+5+...) - 1/2(1+2+3+4+...)
= (1-1/2)(1+2+3+...)
= 1/2*(2-1)(1+2+3+...)
= 1/2*( (2+4+6+8+...) - (1+2+3+4+5+...) )
= 1/2*(-A)

But you said earlier that A is obviously equal to [itex]\infty[/itex], so then now I've manipulated it to be equal to [itex]-\infty[/itex], but at the same time I've shown that A= -1/2*A so does that mean A=0? It seems to have multiple answers, all at the same time.

Infinities, don't mess with them

 

[Routaran]

I think I am starting to get a better idea. Are you saying that the order in which I perform arithmetic matters when I am dealing with infinities and that I cannot just multiply or add any which way I want?
if so, is there a right/wrong way to deal with situations like this, something analogous to BODMAS for example, that guides on exactly how to proceed?

 

[coolul007]

Also, you are claiming after multiplying by (2-1) you have 2 sets that have a different number of elements, one set has one more member, a -1, where did it come from?

 

[Routaran]

Also, you are claiming after multiplying by (2-1) you have 2 sets that have a different number of elements, one set has one more member, a -1, where did it come from?

i see what i was doing incorrect now in that first example
when i said
(2+4+8+16+32+...) - (1+2+4+8+16+...)
i was incorrect to subtract the 2nd number in the 2nd term from the 1st number in the first term. what i should have been doing was 1st number from 1st term - 1st number from 2nd term

so i get
(2+4+8+16+32+...) - (1+2+4+8+16+...)
= (2-1) + (4-2) + (8-4) + (16-8) + ...
= 1+2+4+8+...
and the equality is preserved.

I was very surprised to learn that the order, even in addition, matters.
Thank you for the education.

 

[micromass]

I was very surprised to learn that the order, even in addition, matters.

Yes, that is surprising. But one can say that the order matters here because [itex]1+2+4+8+...[/itex] sums to infinity. But even if the sum was not infinity, even then the order matters. For example, [itex]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...[/itex] is an infinite sum which can be proven to equal a finite number, but still order matters.

Series for which order does not matter are called commutatively convergent series.

 

[Millennial]

To be honest, the sum still equals -1 in a specific sense. In mathematics, this is often called analytic continuation.

The Taylor series of the function [itex]\dfrac{1}{1-x}[/itex] is exactly the geometric series where the powers of x are taken, starting from 0 (when the series exist.) This means that the function we gave analytically continues the geometric series. Even where the series is divergent, a value can be assigned to the sum using the analytic continuation. Note that this DOES NOT equal the value of the actual sum, but in numerous physical and mathematical applications; the analytic continuation becomes important.

The series you gave has a common ratio of 2, hence x=2. Plugging that into the analytic continuation function yields -1, which means -1 can be treated as a finite value that can be assigned to the sum (it does not equal the sum.) In fact, the geometric series which only makes sense when [itex]|x|<1[/itex] is mapped to a meromorphic function with a simple pole at x=1 with residue -1.

 

[cauchyapple]

This is why math people like to define what they are talking about :)

What you've given is basically a special-case proof of the infinite geometric series formula: a_0/1-r. Here, a_0 = 1, and r = 2 so we get -1. Of course, as you noticed, it can't apply here.

Consider the case A = 1+1+1+1...
A = 0+A = 0+1+1+1...

Subtracting "term by term", we will get 0 = 1! Complete nonsense...

This leads us to define infinite series as "limits of partial sums"

Now suppose we have a series a_0+a_1+...

We let A_0 = a_0, A_1 = a_0+a_1 etc... and supposing that the sequences A_n "settles" to some number (converges is the technical word), we let A = lim A_n as n->infinity.

Let us define B in the same way. Now it turns out we CAN do term by term subtraction and arrive at A-B.

Our subtracted series will end up being: (a_0-b_0) + (a_1-b_0) ... which defined in the same partial sums way, is lim (A_n-B_n) as n-> infinity. Since {A_n} and {B_n} are convergent sequences, this evaluates to A-B. Anyways I am sure that this was hard to understand since I didn't give any details, but it's part of the motivation for studying analysis.

 

[Byron Chen]

I wasnt sure which math forum to ask this in so I am putting it here, apologies if its supposed to be elsewhere.

Anyway, my question has to do with an infinite sum.
Suppose we have a series:
1+2+4+8+16+32+...
each term in the series is double the previous term starting with 1

if we multiply this by the number 1 it will remain the same
1+2+4+8+16+32+... = 1(1+2+4+8+16+32+...)

also (2-1) = 1
so we can say
1+2+4+8+16+32+... = (2-1)(1+2+4+8+16+32+...)
if we expand the RHS we get
1+2+4+8+16+32+... = (2+4+8+16+32+64+...) - (1+2+4+8+16+32+...)
1+2+4+8+16+32+... = (2+4+8+16+32+64+...) + (-1-2-4-8-16-32-...)

the first part of the RHS has a set of all positive even numbers and the 2nd part of the set has a set of all negative even numbers
all the even numbers will subtract to 0 and I'm left with
1+2+4+8+16+32+... = -1

what have I done wrong? I can't figure what and where the error is.

I think your misconception is that you treated infinity like an everyday real (or more generally complex) number. I know this may not look intuitive, but ∞-∞ is not 0, but rather it is an indeterminate form! And ∞+1=∞. So what you did above is rather like taking ∞-∞, which I've just told you is not as simple as 0.

Other indeterminate forms are like 0/0, 0 times ∞, and ∞/∞. About the infinity + 1 thing I mentioned above, there is something like: if you have infinite number of hotel rooms and infinite people such that all rooms are occupied, and one more person comes along and wants a room. Solution: Get each occupant to move one room down, and surprisingly enough, nobody would be left without a room. (This does not consider you need infinite amount of time to get your message across)