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 .
If we sum up the second column, we get .
If we sum up the third column, we get .
This pattern continues, hence the sum of all the terms is
Now, let's sum up the rows using partial fractions to get a telescoping series.
If we sum up the first row, we get .
If we sum up the second row, we get .
If we sum up the third row, we get .
This pattern continues, hence the sum of all the terms is
Where did the come from?
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
Here's an even simpler example to show why this is flawed: consider the sum S=1+1+1+⋯ What happens when we group it as follows: S=1+(1+1+1+⋯) That means S=1+S so 0=1. something is obviously wrong here.
Log in to reply
Oh...... Last year there was this note where someone proved that 0=1.... I think I still get it......
Log in to reply
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 ∞=∞+1. We cannot conclude that 0=1, because we do not have additive inverse of infinity, namely that ∞−∞=0.
Of course, conversely, the above is (can be modified to) a proof by contradiction that the sum of the harmonic series is infinity.
Log in to reply
S=1+21+31+⋯ 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.
SupposeThus, S is infinite.
Both series diverge so technically they are both positive infinity and are "equal".
Actually, you are not exactly dealing with a convergent series.
It's because infinity is weird... :)
WHEN WE SUM UP THE FIRST ROW WE WILL GET 1
wait S isn't even well defined
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.
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