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?

## Comments

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TopNewestHere'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 · 2 years, 6 months ago

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– Jeremy Bansil · 2 years, 6 months ago

Oh...... Last year there was this note where someone proved that \(0 = 1\).... I think I still get it......Log in to reply

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 · 2 years, 6 months ago

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Thus, \(S\) is infinite. – Daniel Liu · 2 years, 6 months ago

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– Jake Lai · 2 years, 6 months ago

Yeah, thought so. Nice little note; divergent series are such pesky things.Log in to reply

Both series diverge so technically they are both positive infinity and are "equal". – Yannick Yao · 2 years, 6 months ago

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Actually, you are not exactly dealing with a convergent series. – Agnishom Chattopadhyay · 2 years, 6 months ago

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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 · 9 months, 3 weeks ago

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wait S isn't even well defined – Bill Huang · 2 years, 3 months ago

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WHEN WE SUM UP THE FIRST ROW WE WILL GET 1 – Arnab Karmakar · 2 years, 4 months ago

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It's because infinity is weird... :) – Bill Huang · 2 years, 6 months 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 · 9 months, 3 weeks ago

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