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

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

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

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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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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.

Daniel Liu - 4 years, 10 months ago

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

Jeremy Bansil - 4 years, 10 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.

Calvin Lin Staff - 4 years, 10 months ago

@Calvin Lin – 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.

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

Jake Lai - 4 years, 10 months ago

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

Yannick Yao - 4 years, 10 months ago

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

Agnishom Chattopadhyay Staff - 4 years, 10 months ago

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

Bill Huang - 4 years, 10 months ago

WHEN WE SUM UP THE FIRST ROW WE WILL GET 1

Arnab Karmakar - 4 years, 8 months ago

wait S isn't even well defined

Bill Huang - 4 years, 7 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.

Rick Thomas - 3 years, 1 month 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

Terence Coelho - 3 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$