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 Timios August 27th, 2015 06:14 PM

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.

 Azzajazz August 27th, 2015 11:04 PM

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.

 Country Boy August 28th, 2015 06:21 AM

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.)

 v8archie August 28th, 2015 06:42 AM

Quote:
 Originally Posted by Country Boy (Post 334776) 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 $\displaystyle \frac{a}{1- r}$.

 Country Boy August 28th, 2015 11:41 AM

How in the world did that "2" sneak in there?! Yes, of course. Thank you.

 Timios August 28th, 2015 11:49 AM

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.

 topsquark August 28th, 2015 12:32 PM

Quote:
 Originally Posted by Timios (Post 334877) 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

 v8archie August 28th, 2015 12:44 PM

Quote:
 Originally Posted by Timios (Post 334877) 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.

 v8archie August 28th, 2015 02:30 PM

Quote:
 Originally Posted by topsquark (Post 334878) 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

 Timios August 28th, 2015 03:51 PM

Quote:
 Originally Posted by topsquark (Post 334878) 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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