Kempner Series & MMC

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In Madhava Mathematics Competition 2015 (held in January 2015), we were asked to prove the convergence of Kempner Series (first time proved in 1914). Recently I discovered the paper by A. J. Kempner (http://www.jstor.org/stable/2972074), so in this blog post I will state and prove that problem.

The basic idea behind proof is to divide whole series into chunks (finding symmetry) and then construct a converging geometric series which will act as upper bound of Kempner Seires.

A detailed study of this cartoon has been done by Ben Orlin (http://bit.ly/1KD4shF)

A detailed study of this cartoon has been done by Ben Orlin (http://bit.ly/1KD4shF)

Theorem (Kempner Series, 1914). Harmonic Series,\sum_{k=1}^{\infty} \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} +\ldots, converge, if the denominators do not include all natural numbers 1,2,3,\ldots , but only those numbers which do not contain any figure 9.

Proof: Given Series is:
K = \frac{1}{1}+\ldots +\frac{1}{8}+\frac{1}{10}+\ldots +\frac{1}{18}+\frac{1}{20} + \ldots + \frac{1}{28}+\frac{1}{30}+\ldots +\frac{1}{38}+\frac{1}{40}+\ldots +\frac{1}{48}+\frac{1}{50} + \ldots + \frac{1}{58}+\frac{1}{60}+\ldots+\frac{1}{68}+\frac{1}{70}+\ldots +\frac{1}{78}+\frac{1}{80} + \ldots + \frac{1}{88}+\frac{1}{100}+\ldots
Now we can rewrite above series as:
S = a_1 + a_2 + \ldots + a_n + \ldots
where a_n is the sum of all terms in K of denominator d with 10^{n-1} \leq d < 10^{n}.
Observe that, each term of K which forms part of a_n, is less than or equal to 1/10^{n-1}.

Now count the number of terms of K which are contained in a_1, in a_2, \ldots, in a_n. Clearly, a_1, consists of 8 terms, and a_1 < 8 \cdot 1 < 9. In a_2 there are, as is easily seen, less than 9^2 terms of K, and a_2 < (9^2/10). Altogether there are in K less than 9^2 + 9 terms with denominators under 100.

Assume now that we know the number of terms in K which are contained in a_n to be less than 9^n, for n = 1, 2, 3, \ldots, n. Then, because each term of K which is contained in a_n is not greater than 1/10^{n-1}, we have a_n < (9^n/10^{n-1}), and the total number of terms in K with denominators under 10^n is less than 9^n + 9^{n-1} + 9^{n-2} + \ldots + 9^2 + 9.

Now, let’s go for induction. For n = 1 and n = 2 we have verified all this, and we will now show that if it is true for n, then a_{n+1} < (9^{n+1}/10^n). a_{n+1} contains all terms in K of denominator d, 10^n \leq d < 1^{n+1}. This interval for d can be broken up into the nine intervals,  \alpha \cdot 10^n \leq d < (\alpha + 1)10^n, \alpha = 1,2, \ldots, 9. The last interval does not contribute any term to K, the eight remaining intervals contribute each the same number of terms to K, and this is the same as the number of terms contributed by the whole interval 0 < d < 10^n, that is, by assumption, less than 9^n + 9^{n-1} + 9^{n-2} + \ldots + 9^2 + 9.

Therefore, a_{n+1} contains less than 8(9^n + 9^{n-1} + 9^{n-2} +\ldots + 9^2 + 9) < 9^{n+1} terms of K, and each of these terms is less than or equal to 1/10, we have a_{n+1} < (9^{n+1}/10^n).

Hence, S = a_1 + a_2 + a_3 + \ldots < 9 + \frac{9}{10} + \ldots + \frac{9^{n+1}}{10^n}+ \ldots = 90
Thus, S converges, and since, K=S, K also converges.

——————————–
Note: There is nothing special about 9 here, the above method of proof holds unchanged if, instead of 9, any other figure 1, 2, \ldots, 8 is excluded, but not for the figure 0.

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One response »

  1. Pingback: Four Examples | Gaurish4Math

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