2 ms·
A very cute algebraic way to see this is to do arithmetic in base 4. In base 4 the series on the left is: 0.1 + 0.01 + 0.001 + ... = 0.111... (recurring) Mu
by michael_nielsen 17y ago
A very cute algebraic way to see this is to do arithmetic in base 4. In base 4 the series on the left is:
0.1 + 0.01 + 0.001 + ... = 0.111... (recurring)
Multiplying the right hand side by 3 gives 0.3333... = 1, and so the original series must have just been 1/3.
- ftse 17y agoCute. To illustrate the sum of limits, however, I prefer a graphic with sum -> 1. Take an empty circle then add a chunk of shaded semicircle, then add one quarter, then add one eighth... you get the idea.
- eru 17y agoWith this method you won't be able to tell converging sequences from diverging ones. Suppose I used your circle and put in the first hundred terms of the harmonic sequence (divided by 10 or so, to fit in the unit area).
- jonsen 17y agoVery nice indeed, but you can do it in binary, where the series is 0.01 + 0.0001 + 0.000001 + ... = 0.010101010101... Multiplying the right hand side by 2 gives 0.101010101010... 0.010101010101... + 0.101010101010... ------------------- 0.111111111111... = 1 = 11*0.010101010101...
- michael_nielsen 17y agoThe advantage of my proof is that it generalizes to give the sum of 1/n + 1/n.n + ... Just replace 4 everywhere by n, and 3 by n-1: 1/n + 1/n.n + ... = 0.1111... (in base n) Multiply by (n-1) to get: 0.(n-1)(n-1)(n-1)... = 1 So the original sum must have been 1/(n-1). This is not so easy in binary.