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# Where did the 1 come from?

Consider the following table of numbers:

$\large \begin{array} { c | c |c |c |c |c } \, & & & & & \\ \hline & \frac{1}{ 1 \times 2 } & \frac{1}{2 \times 3 } & \frac{1}{3 \times 4} & \frac{1}{4 \times 5} & \ldots \\ \hline & & \frac{1}{2 \times 3 } & \frac{1}{3 \times 4} & \frac{1}{4 \times 5} & \ldots \\ \hline & & & \frac{1}{3 \times 4} & \frac{1}{4 \times 5} & \ldots \\ \hline & & & & \frac{1}{4 \times 5} & \ldots \\ \hline & & & & & \ddots \\ \end{array}$

If we sum up the first column, we get $$\frac{1}{1 \times 2 } = \frac{1}{2}$$.
If we sum up the second column, we get $$\frac{2}{2 \times 3 } = \frac{1}{3}$$.
If we sum up the third column, we get $$\frac{3}{3 \times 4 } = \frac{1}{4}$$.
This pattern continues, hence the sum of all the terms is

$S = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots$

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

$S = \color{red}{1} + \frac{1}{2} + \frac{1}{3} + \ldots$

Where did the $$\color{red} { 1}$$ come from?

Note by Calvin Lin
1 year, 9 months ago

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Here's an even simpler example to show why this is flawed: consider the sum $S=1+1+1+\cdots$ What happens when we group it as follows: $S=1+(1+1+1+\cdots)$ That means $S=1+S$ so $$0=1$$. something is obviously wrong here. · 1 year, 9 months ago

Oh...... Last year there was this note where someone proved that $$0 = 1$$.... I think I still get it...... · 1 year, 9 months ago

The trick is that the sum of all the terms $$S$$ 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 $$\infty = \infty + 1$$. We cannot conclude that $$0 = 1$$, because we do not have additive inverse of infinity, namely that $$\infty - \infty \neq 0$$.

Of course, conversely, the above is (can be modified to) a proof by contradiction that the sum of the harmonic series is infinity. Staff · 1 year, 9 months ago

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

Thus, $$S$$ is infinite. · 1 year, 9 months ago

Yeah, thought so. Nice little note; divergent series are such pesky things. · 1 year, 9 months ago

Both series diverge so technically they are both positive infinity and are "equal". · 1 year, 9 months ago

Actually, you are not exactly dealing with a convergent series. · 1 year, 9 months ago

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. · 3 weeks, 5 days ago

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 · 3 weeks, 5 days ago

wait S isn't even well defined · 1 year, 6 months ago

WHEN WE SUM UP THE FIRST ROW WE WILL GET 1 · 1 year, 7 months ago