It is well known that the harmonic series, 1/1 + 1/2 + 1/3 + 1/4 + ... , diverges. Consider a *depleted* harmonic series; see below; which contains only terms whose denominator does not contain a 9. (In decimal representation.) Does this series diverge or converge?

S = 1/1 + 1/2 + ... + 1/8 + 1/10 + ... + 1/18 + 1/20 + ... + 1/88 + 1/100 + 1/101 + ...

Grouping terms according to the number of digits in their denominator, we have

S_{n} = (1/1 + ... + 1/8) + (1/10 + ... + 1/18 + 1/20 + ... + 1/88) + (1/100 + ... + 1/888) + ... + (1/10^{n−1} + ... + 1/8...8 [n eights])

Now form another series, which is term-for-term greater than or equal to S_{n}, by setting each term with k digits in the denominator to 1/10^{k−1}:

T_{n} = (1/1 + ... + 1/1) + (1/10 + ... + 1/10) + (1/100 + ... + 1/100) + ... + (1/10^{n−1} + ... + 1/10^{n−1})

Note that the number of k-digit numbers without a 9 is 8×9^{k−1}.

(There are 8 choices for the leading digit, and 9 choices for each of the other digits.)

Hence

T_{n} = 8×1 + (8×9)/10 + (8×9^{2})/10^{2} + ... + (8×9^{n−1})/10^{n−1}

This is a geometric series with first term 8, common ratio 9/10, and n terms.

Therefore T_{n} = 80[1 − (9/10)^{n}]

As n , T_{n} 80.

That is, the infinite series, T, converges to 80; T = T_{∞} = 80.

By the Comparison Test, and since at least one term in S_{n} is strictly less than the corresponding term in T_{n}, the infinite series,

S = (1/1 + ... + 1/8) + (1/10 + ... + 1/18 + 1/20 + ... + 1/88) + (1/100 + ... + 1/888) + ... < 80.

That is, S converges, to a value less than 80.

Spot the flaw in the following two 'proofs' that the harmonic series *converges!*

Let S_{n} denote the sum of those terms of the harmonic series whose denominators do not contain the decimal digit *n*.

We have already seen that S_{9} < 80. It can similarly be shown that S_{n} < 80, for 1 ≤ n ≤ 8, and that S_{0} < 90.

Since each S_{n} is absolutely convergent, S_{0} + S_{1} + ... + S_{9} < 810.

Since each term in the harmonic series occurs in at least one series, S_{n}, it follows that the harmonic series itself converges.

Consider S_{9}, and its complementary series, T_{9}, in which each term contains at least one 9:

S_{9} = 1/1 + 1/2 + ... + 1/8 + 1/10 + ... + 1/18 + 1/20 + ...

T_{9} = 1/9 + 1/19 + ... + 1/89 + 1/90 + ... + 1/99 + 1/109 + ...

Since T_{9} is less, term-for-term, than S_{9}, which converges, it follows from the Comparison Test that T_{9} also converges.

Moreover, both S_{9} and T_{9} are absolutely convergent; hence their sum, the harmonic series, converges.

Depleted harmonic series were first studied by A. J. Kempner in 1914. Consider S_{n}, as defined above. R. Baillie has provided a method for summing the series with great accuracy and economy, given to five decimal places in the table below. (Source: Section 3.3 of Gamma: Exploring Euler's Constant, by Julian Havil.)

Missing digit | Sum |
---|---|

1 | 16.17696 |

2 | 19.25735 |

3 | 20.56987 |

4 | 21.32746 |

5 | 21.83460 |

6 | 22.20559 |

7 | 22.49347 |

8 | 22.72636 |

9 | 22.92067 |

0 | 23.10344 |

- Kempner Series
- The Harmonic Numbers and Series
- In perfect harmony
- Rearranging the Alternating Harmonic Series
- Random Harmonic Series

Source: A. J. Kempner