Where did the 1 come from?

Consider the following table of numbers:

11×212×313×414×512×313×414×513×414×514×5 \large \begin{array} { c | c |c |c |c |c } \, & & & & & \\ \hline & \frac{1}{ 1 \times 2 } & \frac{1}{2 \times 3 } & \frac{1}{3 \times 4} & \frac{1}{4 \times 5} & \ldots \\ \hline & & \frac{1}{2 \times 3 } & \frac{1}{3 \times 4} & \frac{1}{4 \times 5} & \ldots \\ \hline & & & \frac{1}{3 \times 4} & \frac{1}{4 \times 5} & \ldots \\ \hline & & & & \frac{1}{4 \times 5} & \ldots \\ \hline & & & & & \ddots \\ \end{array}

If we sum up the first column, we get 11×2=12 \frac{1}{1 \times 2 } = \frac{1}{2} .
If we sum up the second column, we get 22×3=13 \frac{2}{2 \times 3 } = \frac{1}{3} .
If we sum up the third column, we get 33×4=14 \frac{3}{3 \times 4 } = \frac{1}{4} .
This pattern continues, hence the sum of all the terms is

S=12+13+14+ S = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots

Now, let's sum up the rows using partial fractions to get a telescoping series.
If we sum up the first row, we get i=11i(i+1)=1 \displaystyle \sum_{i=1}^\infty \frac{ 1}{ i (i+1) } = 1 .
If we sum up the second row, we get i=21i(i+1)=12 \displaystyle \sum_{i=2}^\infty \frac{ 1}{ i (i+1) } = \frac{1}{2} .
If we sum up the third row, we get i=31i(i+1)=13 \displaystyle \sum_{i=3}^\infty \frac{ 1}{ i (i+1) } = \frac{1}{3} .
This pattern continues, hence the sum of all the terms is

S=1+12+13+ S = \color{#D61F06}{1} + \frac{1}{2} + \frac{1}{3} + \ldots

Where did the 1 \color{#D61F06} { 1} come from?

Note by Calvin Lin
4 years, 10 months ago

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1 vote

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I am fairy certain the top row does not approach 1... More like it approaches 0.75, The second row does not approach 1/2, more like it approaches 0.2 (both of these are quick evaluations and would require more time to evaluate) but I think the math holds up, just not the assumptions.

Rick Thomas - 3 years, 1 month ago

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Step 1: Assume: 1/2+1/3+1/4+... converges: Then: [insert original post here] Which is a contradiction. Thus, 1/2+1/3+... diverges

Terence Coelho - 3 years, 1 month ago

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wait S isn't even well defined

Bill Huang - 4 years, 7 months ago

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WHEN WE SUM UP THE FIRST ROW WE WILL GET 1

Arnab Karmakar - 4 years, 8 months ago

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It's because infinity is weird... :)

Bill Huang - 4 years, 10 months ago

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Here's an even simpler example to show why this is flawed: consider the sum S=1+1+1+S=1+1+1+\cdots What happens when we group it as follows: S=1+(1+1+1+)S=1+(1+1+1+\cdots) That means S=1+SS=1+S so 0=10=1. something is obviously wrong here.

Daniel Liu - 4 years, 10 months ago

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Oh...... Last year there was this note where someone proved that 0=10 = 1.... I think I still get it......

Jeremy Bansil - 4 years, 10 months ago

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The trick is that the sum of all the terms SS is actually infinite. We can recognize this because we know that the sum of the harmonic series is infinity.

Thus, we are actually saying that =+1 \infty = \infty + 1 . We cannot conclude that 0=1 0 = 1 , because we do not have additive inverse of infinity, namely that 0 \infty - \infty \neq 0 .


Of course, conversely, the above is (can be modified to) a proof by contradiction that the sum of the harmonic series is infinity.

Calvin Lin Staff - 4 years, 10 months ago

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@Calvin Lin Yeah, thought so. Nice little note; divergent series are such pesky things.

Jake Lai - 4 years, 10 months ago

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@Calvin Lin Suppose S=1+12+13+S=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots is finite. Then, by the summations we did above, S1=SS-1 = S. Since SS is finite, we can subtract SS from both sides to get 1=0-1=0, contradiction.

Thus, SS is infinite.

Daniel Liu - 4 years, 10 months ago

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Both series diverge so technically they are both positive infinity and are "equal".

Yannick Yao - 4 years, 10 months ago

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Actually, you are not exactly dealing with a convergent series.

Agnishom Chattopadhyay Staff - 4 years, 10 months ago

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