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?
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Top NewestHere'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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Oh...... Last year there was this note where someone proved that \(0 = 1\).... I think I still get it...... – Jeremy Bansil · 2 years, 6 months ago
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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 · 2 years, 6 months ago
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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. – Daniel Liu · 2 years, 6 months ago
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Yeah, thought so. Nice little note; divergent series are such pesky things. – Jake Lai · 2 years, 6 months ago
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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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