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August 27th, 2015, 06:14 PM   #1
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Proof that an Infinite Series Adds to 1

We will prove that:
$\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}... = 1$ (exactly)

Proof:
(1) Let n = $\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$

Multiply each side of (1) by 2

(2) 2n = $\displaystyle 1 + \frac{1}{2}+\frac{1}{4}+\frac{1}{8}...$

Subtract each side of line (1) from each side of line (2)

(3) n = 1

One can also use this technique to prove that .999... is exactly equal to 1.
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August 27th, 2015, 11:04 PM   #2
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Be careful when using proofs like this.
Saying "Let $n = \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + ...$" implies that the series converges, which may not be the case.

There are (false) proofs that the series $1 - 1 + 1 - 1 + ...$ converges to $0$, $1$ and $\dfrac{1}{2}$ using this principle, but the series does not converge at all.

If you want to formalise this proof, you have to first prove that the series converges.
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August 28th, 2015, 06:21 AM   #3
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This is, in fact, a "geometric series". A geometric series is any series of the form $\displaystyle \sum_{n=0}^\infty ar^n= a+ ar+ ar^2+ ar^3+ \cdot\cdot\cdot$.

One can prove that a geometric series converges if and only if |r|< 1 and then converges to $\displaystyle \frac{a}{1- r^2}$.

Here, we can write the series as $\displaystyle \frac{1}{2}\left(1+ \frac{1}{2}+ \frac{1/4}+ \cdot\cdot\cdot\right)$ so 1/2 times a geometric sequence with a= 1 and r= 1/2. Since 1/2< 1, this converges to $\displaystyle \frac{1}{2}\frac{1}{1- \frac{1}{2}}= \frac{1}{2}\frac{2}{2- 1}= 1$.

(This is also a geometric series with a= 1/2 and r= 1/2 so converges to $\displaystyle \frac{\frac{1}{2}}{1- \frac{1}{2}}= \frac{\frac{1}{2}}{\frac{1}{2}}= 1$)

(It can also be thought of as a geometric series with a= 1 and r= 1/2 with the first "a" term missing. The geometric series with a= 1 and r= 1/2, $\displaystyle 1+ \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+ \cdot\cdot\cdot$ converges to $\displaystyle \frac{1}{1- \frac{1}{2}}= 2$ and subtracting the missing first term, "1" gives a sum of 1.)
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August 28th, 2015, 06:42 AM   #4
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Originally Posted by Country Boy View Post
One can prove that a geometric series converges if and only if |r|< 1 and then converges to $\displaystyle \frac{a}{1- r^2}$.
That should be $\displaystyle \frac{a}{1- r}$.
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August 28th, 2015, 11:41 AM   #5
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How in the world did that "2" sneak in there?! Yes, of course. Thank you.
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August 28th, 2015, 11:49 AM   #6
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Perhaps I'm stupid, but I see no necessity of first proving that the series converges. Whether it converges or not, the proof I provided is purely algebraic:

Multiplying each side of an equation by 2, and
Subtracting the same thing from each side of an equation.
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August 28th, 2015, 12:32 PM   #7
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Quote:
Originally Posted by Timios View Post
Perhaps I'm stupid, but I see no necessity of first proving that the series converges. Whether it converges or not, the proof I provided is purely algebraic:

Multiplying each side of an equation by 2, and
Subtracting the same thing from each side of an equation.
If the sum of the series does not converge what does the "value" of the sum mean in the first place?

For example:
S = 1 + 2 + 3 + 4 + ...

2S = 2 + 4 + 6 + 8 + ....

S = 2S - S = -1 - 3 - 5 - 7 - ...

But can we say that 1 + 2 + 3 + 4 + ... = -1 - 3 - 5 - 7 - .... ?

-Dan
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August 28th, 2015, 12:44 PM   #8
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Originally Posted by Timios View Post
Perhaps I'm stupid, but I see no necessity of first proving that the series converges.
The example you were given is as follows.

Suppose that $$s = 1 - 1 + 1 - 1 + \cdots$$Then $$-s = -1 + 1 - 1 + 1 \cdots = -1 + s$$Thus $$2s = 1 \implies \boxed{s = \tfrac12}$$

This uses all the same techniques that yours does, but is patently not true. The partial sums never get closer to $\frac12$ at all!

What you are really saying is that "I get the right answer, so why should I prove that the method is sound". But that means that what you have is not a proof that the sum is what you say it is because applied to another sequence we don't know whether the result is correct or not. We have to check by another method.
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Last edited by v8archie; August 28th, 2015 at 12:48 PM.
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August 28th, 2015, 02:30 PM   #9
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Originally Posted by topsquark View Post
For example:
S = 1 + 2 + 3 + 4 + ...
2S = 2 + 4 + 6 + 8 + ....
Doing the subtraction as Timios did in the original post we get
S = 1 + (2 - 2) + (3 - 4) + (4 - 6) + ...
S = 1 + 0 - 1 - 2 - 3 - ...
S = 1 - S
S = 1/2
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August 28th, 2015, 03:51 PM   #10
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Quote:
Originally Posted by topsquark View Post
If the sum of the series does not converge what does the "value" of the sum mean in the first place?

For example:
S = 1 + 2 + 3 + 4 + ...

2S = 2 + 4 + 6 + 8 + ....

S = 2S - S = -1 - 3 - 5 - 7 - ...

But can we say that 1 + 2 + 3 + 4 + ... = -1 - 3 - 5 - 7 - .... ?

-Dan
I don't know what you are doing here. But if you do the analagous thing that I did, using your S and 2S, you would calculate 2S-S as follows:

2S= 2 + 4 + 6 + 8 + ...
S= 1 + 2 + 3 + 4 + ...

Subtract both sides and you have:

S = 1 + 2 + 3 + 4 + ...
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