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Solution to puzzle 72: Depleted harmonic series

Skip restatement of puzzle.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

Sn = (1/1 + ... + 1/8) + (1/10 + ... + 1/18 + 1/20 + ... + 1/88) + (1/100 + ... + 1/888) + ... + (1/10n−1 + ... + 1/8...8 [n eights])

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

Tn = (1/1 + ... + 1/1) + (1/10 + ... + 1/10) + (1/100 + ... + 1/100) + ... + (1/10n−1 + ... + 1/10n−1)

Note that the number of k-digit numbers without a 9 is 8×9k−1.
(There are 8 choices for the leading digit, and 9 choices for each of the other digits.)

Hence

Tn = 8×1 + (8×9)/10 + (8×92)/102 + ... + (8×9n−1)/10n−1

This is a geometric series with first term 8, common ratio 9/10, and n terms.
Therefore Tn = 80[1 − (9/10)n]

As n tends to infinity, Tn tends to 80.
That is, the infinite series, T, converges to 80; T = T = 80.

By the Comparison Test, and since at least one term in Sn is strictly less than the corresponding term in Tn, 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.


Remarks

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

Proof by Sum

Let Sn denote the sum of those terms of the harmonic series whose denominators do not contain the decimal digit n.
We have already seen that S9 < 80.  It can similarly be shown that Sn < 80, for 1 ≤ n ≤ 8, and that S0 < 90.

Since each Sn is absolutely convergent, S0 + S1 + ... + S9 < 810.
Since each term in the harmonic series occurs in at least one series, Sn, it follows that the harmonic series itself converges.

Proof by Comparison

Consider S9, and its complementary series, T9, in which each term contains at least one 9:

S9 = 1/1 + 1/2 + ... + 1/8 + 1/10 + ... + 1/18 + 1/20 + ...

T9 = 1/9 + 1/19 + ... + 1/89 + 1/90 + ... + 1/99 + 1/109 + ...

Since T9 is less, term-for-term, than S9, which converges, it follows from the Comparison Test that T9 also converges.
Moreover, both S9 and T9 are absolutely convergent; hence their sum, the harmonic series, converges.

The Kempner-depleted harmonic sums

Depleted harmonic series were first studied by A. J. Kempner in 1914.  Consider Sn, 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.)

Kempner-depleted harmonic sums
Missing digitSum
116.17696
219.25735
320.56987
421.32746
521.83460
622.20559
722.49347
822.72636
922.92067
023.10344

Further reading

  1. Kempner Series
  2. The Harmonic Numbers and Series
  3. In perfect harmony
  4. Rearranging the Alternating Harmonic Series
  5. (Adobe) Portable Document Format Random Harmonic Series

Source: A. J. Kempner

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