CTK Insights

27 Oct

That Divergent Harmonic Series

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The harmonic series, i.e., the series of the reciprocals of integers is a prototypical divergent series: the sums

\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}

grow, albeit very slowly, without bound. There is no finite value S that could be ascribed as its sum. Assuming that such a value exists leads to a contradiction. A contradiction could be achieved in a multitude of ways. R. P. Boas once listed five of them, but there are more. Several has been listed by R. Honsberger and by W. Dunham. Kifowit and Stamps gave 20 proofs. More recently, the harmonic series attracted bloggers, at Let's Play Math and Random Walks.

Below I'll give a few proofs.

So assume that the series is convergent, i.e., for a finite S,

S =  \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} + \cdots.

Let's mention the obvious, viz., that all the terms in the harmonic series are positive. It is well known that, for such series, the terms in the sum could be shuffled in any order without affecting the sum.

Proof 1 (Johann Bernoulli)

S =  \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{1} + \frac{1}{3} + \cdots.
S =  \frac{1}{2}(\frac{1}{1} + \frac{1}{2} + \cdots) + (\frac{1}{1} + \frac{1}{3} + \cdots).
S =  \frac{1}{2}S + (\frac{1}{1} + \frac{1}{3} + \cdots).
\frac{1}{2}S = \frac{1}{2} + \frac{1}{4} + \cdots = \frac{1}{1} + \frac{1}{3} + \cdots,

which is absurd as the sum of reciprocals of the even and odd integers could not be equal. To see that compare then term by term.

Proof 2 (Johann Bernoulli)

S - 1 = \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \cdots
S - 1 = \frac{1\times 1}{1\times 2} + \frac{2\times 1}{2\times 3} + \cdots + \frac{({n}-{1})\times 1}{({n}-{1})\times n} + \cdots
S - 1 =
    (\frac{1}{1\times 2} + \frac{1}{2\times 3} + \frac{1}{3\times 4} + \cdots) +
        (\frac{1}{2\times 3} + \frac{1}{3\times 4} + \frac{1}{4\times 5} + \cdots) +
                (\frac{1}{3\times 4} + \frac{1}{4\times 5} + \cdots) + \cdots

Now observe that every row in that infinite sum of infinite sums is a telescoping series. The first row sums to 1, the second to \frac{1}{2}, the third to \frac{1}{3}, and so on. We get

S - 1 = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots = S,

making 2S = 1, i.e., S = \frac{1}{2} whereas already the first term in the series is 1. A contradiction.

Proof 3

Combine terms in pairs and, in every pair, replace the largest term with the smallest:

S =  \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots > (\frac{1}{2}+\frac{1}{2})+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{6} +\frac{1}{6})+\cdots.

S =  \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots > {2} \times (\frac{1}{2}+\frac{1}{4}+\frac{1}{6})+\cdots.

S =  \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots > \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots = S.

That is S > S. A contradiction.

Proof 3'

The terms in the series can be combined m, m > 1, at a time with exactly the same result.

Proof 4 (Nicole Oresme)

The terms of the series could be combined into growing groups of 2^{k} terms, k=1,2,3,\cdots:

S =  \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots >

        \frac{1}{1}+\frac{1}{2}+{2} \times \frac{1}{4}+{4} \times \frac{1}{8}+\cdots

        \frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots

which clearly grows without bound.

Proof 5

There are 9 one-digit numbers; their reciprocals exceed \frac{1}{10} each. The sum of the reciprocals exceeds \frac{9}{10}. There are 90 2-digit numbers; their reciprocals exceed \frac{1}{100} each. The sum of the reciprocals of the 2-digit numbers exceeds \frac{90}{100} = \frac{9}{10}. Similarly, the sum of the reciprocals of all 3-digit numbers exceeds \frac{900}{1000} = \frac{9}{10}, and so on. The sum of all the reciprocals of integers (i.e. the sum of the harmonic series) exceeds k \times \frac{9}{10}, for any positive integer k.

Proof 6 (Pietro Mengolli)

The proof is based on the observation that, for a > 1,

\frac{1}{a-1} + \frac{1}{a} + \frac{1}{a+1} > \frac{3}{a}.

This suggests grouping the terms of the harmonic series by 3 starting with second one. The process leads to the inequality S > 1 + 3S, which appears to imply that S < - \frac{1}{2}.

The required inequality follows from \frac{1}{n-1}+\frac{1}{n+1} = \frac{2n}{n^{2}-1} > \frac{2n}{n^2} = \frac{2}{n}.

Proof 7 (A. Cusumano)

S = (\frac{1}{1} + \frac{1}{2}) + (\frac{1}{3} + (\frac{1}{4}) + (\frac{1}{5} + \frac{1}{6}) + \cdots =

     (\frac{1}{1} + \frac{1}{2}) + (\frac{1}{2} + \frac{1}{12}) + (\frac{1}{3} + \frac{1}{30}) + \cdots =
     S + (\frac{1}{2} + \frac{1}{12} + \frac{1}{30} + \cdots).

A contradiction.

References

  1. R. Honsberger, Mathematical Plums, MAA, 1979
  2. R. Honsberger, Mathematical Gems II, MAA, 1976
  3. W. Dunham, Journey through Genius, Penguin Books, 1991
  4. A. Cusumano, The harmonic series diverges, American Mathematical Monthly 105(7), 608.
  5. S. J. Kifowit, T. A. Stamps, The Harmonic Series Diverges Again and Again, The AMATYC Review, Vol. 27, No. 2, Spring 2006, pp. 31–43

One Response to “That Divergent Harmonic Series”

  1. 1
    sahil Says:

    please send me answer of these questions
    1>using mathematical induction to prove that
    1+2+2^2+2^3+2^4+......+=2^(n+1)-1
    2>using mathematical induction prove that
    1^2+2^2+3^2+4^2+........+n^2=n/6(n+1)(2n+1)
    please send me its soluation
    thanks

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