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michael_nielsen

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.

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Very 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...

The 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.

Cute. 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.

With 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).

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