# Solution to puzzle 72: Depleted harmonic series

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 , Tn 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